Q: Question 5The correct form of Newton's second law is —F = ΔpΔt\dfrac{Δ\text p}{Δ\text t}ΔtΔpF = m ΔvΔt\dfrac{Δ\text v}{Δ\text t}ΔtΔvF = v ΔmΔt\dfrac{Δ\text m}{Δ\text t}ΔtΔmF = mv

F = ΔpΔt\dfrac{Δ\text p}{Δ\text t}ΔtΔpReason — As we know,Force = Rate of change of momentumF = ΔpΔt\dfrac{Δ\text p}{Δ\text t}ΔtΔp= Δ(mv)Δt\dfrac{Δ\text {(mv)}}{Δ\text t}ΔtΔ(mv)= m(Δv)Δt\dfrac{\text m(Δ\text v)}{Δ\text t}Δtm(Δv)= maThus, Newton’s second law will be F = ΔpΔt\dfrac{Δ\text p}{Δ\text t}ΔtΔp

Q: Question 12A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied.

As we know,S = ut + (12\dfrac{1}{2}21) at2Given,m = 500 gConverting g to kg,1000 g = 1 kg500 g = (11000\dfrac{1}{1000}10001) x 500 kg100 g = 0.5 KgHence,m = 0.5 Kgu = 0s = 4 mt = 2 sSubstituting the values in the formula above we get,4=(0×2)+(12×a×22)4=0+(12×a×4)4=2a⇒a=42⇒a=2 m s−24 = (0 \times 2) + (\dfrac{1}{2} \times \text a \times 2^2) \\[0.5em] 4 = 0 + (\dfrac{1}{2} \times \text a \times 4) \\[0.5em] 4 = 2\text a \\[0.5em] \Rightarrow \text a = \dfrac{4}{2} \\[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \\[0.5em]4=(0×2)+(21×a×22)4=0+(21×a×4)4=2a⇒a=24⇒a=2 m s−2Hence,a = 2 m s -2Now,Force (f) = mass (m) x acceleration (a)Substituting the values in the formula above we get,F=0.5×2⇒F=1 N\text F = 0.5 \times 2 \\[0.5em] \Rightarrow \text F = 1\ \text N \\[0.5em]F=0.5×2⇒F=1 NHence,Magnitude of force = 1 N

Question 1

Question 5The correct form of Newton's second law is —F = ΔpΔt\dfrac{Δ	ext p}{Δ	ext t}ΔtΔp​F = m ΔvΔt\dfrac{Δ	ext v}{Δ	ext t}ΔtΔv​F = v ΔmΔt\dfrac{Δ	ext m}{Δ	ext t}ΔtΔm​F = mv

Accepted Answer

F = ΔpΔt\dfrac{Δ	ext p}{Δ	ext t}ΔtΔp​Reason — As we know,Force = Rate of change of momentumF = ΔpΔt\dfrac{Δ	ext p}{Δ	ext t}ΔtΔp​= Δ(mv)Δt\dfrac{Δ	ext {(mv)}}{Δ	ext t}ΔtΔ(mv)​= m(Δv)Δt\dfrac{	ext m(Δ	ext v)}{Δ	ext t}Δtm(Δv)​= maThus, Newton’s second law will be F = ΔpΔt\dfrac{Δ	ext p}{Δ	ext t}ΔtΔp​

Question 2

Question 2The linear momentum of a ball of mass 50 g is 0.5 kg m s-1. Find it's velocity.

Accepted Answer

As we know,linear momentum (p) = mass (m) x velocity (v)Given,p = 0.5 kg m s-1m = 50 gConverting g to kg,1000 g=1 kg50 g=(11000)×50⇒50 g=0.05 kg1000\ \text g = 1\ \text {kg} \[0.5em] 50\ \text g = (\dfrac{1}{1000}) \times 50 \[0.5em] \Rightarrow 50\ \text g = 0.05\ \text {kg} \[0.5em]1000 g=1 kg50 g=(10001​)×50⇒50 g=0.05 kgHence, m = 0.05 kgSubstituting the values in the formula above we get,0.5=0.05×v⇒v=0.50.05⇒v=0.10.01⇒v=10 m s−10.5 = 0.05 \times \text v \[0.5em] \Rightarrow \text v = \dfrac{0.5}{0.05} \[0.5em] \Rightarrow \text v = \dfrac{0.1}{0.01} \[0.5em] \Rightarrow \text v = 10\ \text {m s}^{-1}\[0.5em]0.5=0.05×v⇒v=0.050.5​⇒v=0.010.1​⇒v=10 m s−1Hence, velocity = 10 m s-1

Question 3

Question 6A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate: (i) the velocity acquired by the body, and (ii) change in momentum of the body.

Accepted Answer

As we know,Force (f) = mass (m) x acceleration (a)Given,f = 10 Nm = 2 kgt = 3 su = 0Substituting the values in the formula above we get,10=2×a⇒a=102⇒a=5 m s−210 = 2 \times \text a \[0.5em] \Rightarrow \text a = \dfrac{10}{2} \[0.5em] \Rightarrow \text a = 5\ \text {m s}^{-2} \[0.5em]10=2×a⇒a=210​⇒a=5 m s−2Hence,Acceleration produced by the body = 5 m s-2The 1st equation of motion states that;v = u + atSubstituting the values in the formula above we get,v=0+(5×3)⇒v=15 m s−1\text v = 0 + (5 \times 3) \[0.5em] \Rightarrow v = 15\ \text {m s}^{-1}v=0+(5×3)⇒v=15 m s−1Hence, velocity of the body is 15 m s-1(ii) As we know,Change in momentum = Final momentum - Initial momentumInitial momentum = mu = 2 x 0 = 0Final momentum = mv = 2 x 15 = 30 kg m s-1Substituting the values in the formula above we get,Change in momentum = 30 - 0 = 30 kg m s-1

Question 4

Question 7A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.

Accepted Answer

(i) Velocity (v) = distance (S)time (t)\dfrac{\text {distance (S)}}{\text {time (t)}}time (t)distance (S)​Given,s = 100 m in 5 sSubstituting the values in the formula above we get,v=1005v=20 m s−1\text v = \dfrac{100}{5} \[0.5em] \text v = 20\ \text {m s}^{-1}\[0.5em]v=5100​v=20 m s−1Hence, velocity acquired by the body = 20 m s-1(ii) As we know,v2 - u2 = 2asGiven,s = 100 mu = 0v = 20 m s-1Substituting the values in the formula above we get,202−02=2×a×100400=200×a⇒a=2 m s−220^2 - 0^2 = 2 \times \text a \times 100 \[0.5em] 400 = 200 \times \text a \[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \[0.5em]202−02=2×a×100400=200×a⇒a=2 m s−2Hence, a = 2 m s-2(iii) As we know,Force (f) = mass (m) x acceleration (a)Given,m = 100 kga = 2 m s -2Substituting the values in the formula above we get,F=100×2⇒F=200 N\text F = 100 \times 2 \[0.5em] \Rightarrow \text F = 200\ \text N \[0.5em]F=100×2⇒F=200 NHence,Magnitude of force = 200 N

Question 5

Question 10A cricket ball of mass 150 g moving at a speed of 25 m s-1 is brought to rest by a player in 0.03 s. Find the average force applied by the player.

Accepted Answer

The 1st equation of motion states that;v = u + atGiven,m = 150 gConverting g to kg,1000 g = 1 kg150 g = (11000\dfrac{1}{1000}10001​) x 150150 g = 0.15 kgHence, m = 0.15 kgu = 25 m s-1v = 0t = 0.03 sSubstituting the values in the formula above we get,0=25+(a×0.03)⇒a=−250.03⇒a=−833.33 m s−20 = 25 + (\text a \times 0.03) \[0.5em] \Rightarrow \text a = -\dfrac{25}{0.03} \[0.5em] \Rightarrow \text a = - 833.33\ \text {m s}^{-2} \[0.5em]0=25+(a×0.03)⇒a=−0.0325​⇒a=−833.33 m s−2Hence, acceleration of the ball = - 833.33 m s-2`Now,Force (f) = mass (m) x acceleration (a)Substituting the values in the formula above we get,F=0.15×(−833.33)⇒F=−125 N\text F = 0.15 \times (- 833.33) \[0.5em] \Rightarrow \text F = -125\ \text N \[0.5em]F=0.15×(−833.33)⇒F=−125 NHence,Magnitude of force = -125 N

Question 6

Question 12A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied.

Accepted Answer

As we know,S = ut + (12\dfrac{1}{2}21​) at2Given,m = 500 gConverting g to kg,1000 g = 1 kg500 g = (11000\dfrac{1}{1000}10001​) x 500 kg100 g = 0.5 KgHence,m = 0.5 Kgu = 0s = 4 mt = 2 sSubstituting the values in the formula above we get,4=(0×2)+(12×a×22)4=0+(12×a×4)4=2a⇒a=42⇒a=2 m s−24 = (0 \times 2) + (\dfrac{1}{2} \times \text a \times 2^2) \[0.5em] 4 = 0 + (\dfrac{1}{2} \times \text a \times 4) \[0.5em] 4 = 2\text a \[0.5em] \Rightarrow \text a = \dfrac{4}{2} \[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \[0.5em]4=(0×2)+(21​×a×22)4=0+(21​×a×4)4=2a⇒a=24​⇒a=2 m s−2Hence,a = 2 m s -2Now,Force (f) = mass (m) x acceleration (a)Substituting the values in the formula above we get,F=0.5×2⇒F=1 N\text F = 0.5 \times 2 \[0.5em] \Rightarrow \text F = 1\ \text N \[0.5em]F=0.5×2⇒F=1 NHence,Magnitude of force = 1 N

Question 7

Question 13A car of mass 480 kg moving at a speed of 54 km h-1 is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.

Accepted Answer

We know that,acceleration = v−ut\dfrac{\text v - \text u}{\text t}tv−u​Given,m = 480 kgu = 54 km h-1converting km h-1 to m s-154km h−1=54×(1000m60×60s)54km h−1=54×(10m36s)54km h−1=540m36s54km h−1=15m s−154 \text {km h}^{-1} = 54 \times (\dfrac{1000 \text {m}}{60 \times 60 \text {s}}) \[0.5em] 54 \text {km h}^{-1} = 54 \times (\dfrac{10 \text {m}}{36 \text {s}}) \[0.5em] 54 \text {km h}^{-1} = \dfrac{540 \text {m}}{36 \text {s}} \[0.5em] 54 \text {km h}^{-1} = 15 {\text {m s}^{-1}} \[0.5em]54km h−1=54×(60×60s1000m​)54km h−1=54×(36s10m​)54km h−1=36s540m​54km h−1=15m s−1Hence, u = 15 m s-1v = 0t = 10 sSubstituting the values in the formula above we get,a=0−1510⇒a=−1.5 m s−2\text a = \dfrac{0 - 15}{10} \[0.5em] \Rightarrow \text a = -1.5\ \text{m s}^{-2} \[0.5em]a=100−15​⇒a=−1.5 m s−2Hence, a = - 1.5 m s -2The negative sign shows that it is retardation.Now,Force (f) = mass (m) x acceleration (a)Substituting the values in the formula above we get,F=480×(1.5)⇒F=720 N\text F = 480 \times (1.5) \[0.5em] \Rightarrow \text F = 720\ \text N \[0.5em]F=480×(1.5)⇒F=720 NHence,Magnitude of force applied by the brakes = 720 N

Question 8

Question 14A car is moving with a uniform velocity of 30 m s-1. It is stopped in 2 s by applying a force of 1500 N through it's brakes. Calculate (a) the change in momentum of car, (b) the retardation produced in car, and (c) the mass of the car.

Accepted Answer

(i) We know that,acceleration = v−ut\dfrac{\text v - \text u}{\text t}tv−u​Given,v = 0u = 30 m s-1t = 2 sf = 1500 NSubstituting the values in the formula above we get,a=0−302⇒a=−15 m s−2\text a = \dfrac{0 - 30}{2} \[0.5em] \Rightarrow \text a = - 15\ \text{m s}^{-2} \[0.5em]a=20−30​⇒a=−15 m s−2The negative sign shows that it is retardation. So retardation = 15 m s -2Now,Force (f) = mass (m) x acceleration (a)Substituting the values in the formula above we get,1500=m×(15)⇒m=100 kg1500 = \text m \times (15) \[0.5em] \Rightarrow \text m = 100 \text{ kg} \[0.5em]1500=m×(15)⇒m=100 kgHence,Mass of the car = 100 kgAs we know,Change in momentum = Final momentum - Initial momentumInitial momentum = mu = 100 x 30 = 3000 kg m s-1Final momentum = mv = 100 x 0 = 0Substituting the values in the formula above we get,Change in momentum = 0 - 3000 = - 3000 kg m s-1Hence,(i) change in momentum = 3000 kg m s-1(ii) The retardation produced in the car = 15 m s -2(iii) The mass of the car = 100 Kg

Question 9

Question 3Two bodies A and B of same mass are moving with velocities v and 2v respectively.Compare :(i) their inertia (ii) their momentum.

Accepted Answer

(i) The factor on which inertia of a body depends is mass.More the mass, more is the inertia of the body. Thus, if mass of two bodies is same their inertia will also be same.Hence, inertia of A and B will be same as mass of both A and B are same.Hence, ratio of their inertia is 1 : 1(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)Given, mass of A and B are equal.For A,PA = mvAFor B,PB = mvBRatio between the two is —PAPB=mvAmvB\dfrac{{\text P}\text A}{\text {P}\text B} = \dfrac{\text m{\text v}\text A}{\text m{\text v}\text B} \[0.5em]PBPA=mvBmvAAs, mass of A = mass of B , hence,PAPB=vAvB\dfrac{{\text P}\text A}{{\text P}\text B} = \dfrac{{\text v}\text A}{\text {v}\text B} \[0.5em]PBPA=vBvASubstituting the values, we get,PAPB=v2v⇒PAPB=12\dfrac{{\text P}\text A}{{\text P}\text B} = \dfrac{\text v}{2\text v} \[0.5em] \Rightarrow \dfrac{{\text P}\text A}{\text {P}\text B} = \dfrac{1}{2} \[0.5em]PBPA=2vv⇒PBPA=21Hence, ratio between the momentum of A and B is 1 : 2

Question 10

Question 4Two balls A and B of masses m and 2m are in motion with velocities 2v and v respectively.Compare :(i) their inertia, (ii) their momentum, and (iii) the force needed to stop them in same time.

Accepted Answer

(i) Given,Mass of A = mMass of B = 2mThe factor on which inertia of a body depends is mass.More the mass, more is the inertia of the body.Therefore,Inertia of AInertia of B=mass of Amass of B\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {mass of A}}{\text {mass of B}} \[0.5em]Inertia of BInertia of A=mass of Bmass of ASubstituting the values, we get,Inertia of AInertia of B=m2mInertia of AInertia of B=12\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {m}}{\text {2m}} \[0.5em] \dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {1}}{\text {2}} \[0.5em]Inertia of BInertia of A=2mmInertia of BInertia of A=21Hence, ratio of their inertia = 1 : 2(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)Given,vA = 2vvB = vRatio between the two is —PAPB=(mv)A(mv)B\dfrac{{\text P}\text A}{\text {P}\text B} = \dfrac{\text {(mv)}\text A}{\text {(mv)}\text B} \[0.5em]PBPA=(mv)B(mv)ASubstituting the values, we get,PAPB=m×2v2m×vPAPB=2mv2mv⇒PAPB=11\dfrac{\text {P}\text A}{{\text P}\text B} = \dfrac{\text m \times 2\text v}{2\text m \times \text v} \[0.5em] \dfrac{{\text P}\text A}{\text {P}\text B} = \dfrac{2\text {mv}}{2\text {mv}} \[0.5em] \Rightarrow \dfrac{{\text P}\text A}{{\text P}\text B} = \dfrac{1}{1} \[0.5em]PBPA=2m×vm×2vPBPA=2mv2mv⇒PBPA=11Hence, ratio between the momentum of A and B is 1 : 1(iii) According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it and as the ratio of momentum between A and B is 1 : 1, hence, ratio of force needed to stop A and B is also 1 : 1

Question 11

Question 2(viii)Assertion (A) : If the distance between bodies of masses M1 and M2 is increased by a factor of 4, the gravitational force reduces by 1/16.Reason (R) : The gravitational force is inversely proportional to the square of distance between two bodies.both A and R are true and R is the correct explanation of Aboth A and R are true and R is not the correct explanation of Aassertion is false but reason is trueassertion is true but reason is false

Accepted Answer

both A and R are true and R is the correct explanation of AExplanationAssertion (A) is true because according to Newton’s law of gravitation :F=GM1M2r2	ext F=	ext G\dfrac{	ext M_1	ext M_2}{	ext r^2}F=Gr2M1​M2​​If distance r becomes 4 times, then :F′=F42=F16	ext F'=\dfrac {	ext F}{4^2}=\dfrac {	ext F}{16}F′=42F​=16F​So, the force becomes 1/16th of the original.​Reason (R) is true because this is exactly what Newton’s law states :F∝1r2	ext F\propto\dfrac{1}{	ext r^2}F∝r21​Since the reason correctly explains the assertion, both are true and reason is the correct explanation.

Question 12

Question 1The force of attraction between two bodies at a certain separation is 10 N. What will be the force of attraction between them if the separation is reduced to half ?

Accepted Answer

As we know,F=GMmR2\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}F=GR2Mm​Given,F = 10 N, when separation = R,Hence,F1 = G M mR2\dfrac{\text {G M m}}{\text R^2}R2G M m​ = 10 N     [Equation 1]If separation = R2\dfrac{\text R}{2}2R​ then,Substituting the values in the formula above, we get,F2=G M m(R2)2F2=4G M mR2\text F_2 = \dfrac{\text {G M m}}{\big(\dfrac{\text R}{2}\big)^2} \[0.5em] \text F_2 = \dfrac{4 \text {G M m}}{\text R^2} \[0.5em]F2​=(2R​)2G M m​F2​=R24G M m​F2=4×(G M mR2)\text F_2 = 4 \times \big(\dfrac{\text {G M m}}{\text R^2}\big)F2​=4×(R2G M m​)    [Equation 2]Substituting the value of equation 1 in equation 2 we get,F2 = 4 x 10 N = 40 N

Question 13

Question 9A ball is released from a height and it reaches the ground in 3 s. If g = 9.8 m s-2, find —(a) the height from which the ball was released,(b) the velocity with which the ball will strike the ground.

Accepted Answer

(a) As we know from the equation of motion;s = ut + 12\dfrac{1}{2}21​gt2where, s = heightGiven,t = 3 sg = 9.8 m s-2initial velocity (u) = 0Substituting the values in the formula above, we get,S=(0×t)+(12×9.8×32)S=0+(12×9.8×9)S=44.1 m\text S = (0 \times t) + (\dfrac{1}{2} \times 9.8 \times 3^2) \[0.5em] \text S = 0 + (\dfrac{1}{2} \times 9.8 \times 9) \[0.5em] \text S = 44.1\ \text m \[0.5em]S=(0×t)+(21​×9.8×32)S=0+(21​×9.8×9)S=44.1 mHence, the height from which the ball was released = 44.1 m(b) From the equation of motion,v2 = u2 - 2gswhere, v = final velocitySubstituting the values in the formula, we get,v2=02−(2×9.8×44.1)v2=2×9.8×44.1⇒v2=864.36⇒v=29.4 m s−1\text v^2 = 0^2 - (2 \times 9.8 \times 44.1) \[0.5em] \text v^2 = 2 \times 9.8 \times 44.1 \[0.5em] \Rightarrow \text v^2 = 864.36 \[0.5em] \Rightarrow \text v = 29.4 \text{ m s}^{-1} \[0.5em]v2=02−(2×9.8×44.1)v2=2×9.8×44.1⇒v2=864.36⇒v=29.4 m s−1Hence, velocity with which the ball strikes the ground = 29.4 m s-1

Question 14

Question 11A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m s-2find —(a) the initial velocity of the ball(b) the final velocity of the ball on reaching the ground and(c) the total time of journey of the ball.

Accepted Answer

(a) As we know, from the equation of motion,v2 = u2 - 2gs (a = - g as movement is against gravity )Given,s = 20 mg = 10 m s-2v = 0Substituting the values in the formula, we get,0=u2−2×10×20u2=400⇒u=20 m s−10 = \text u^2 - 2 \times 10 \times 20 \[0.5em] \text u^2 = 400 \[0.5em] \Rightarrow \text u = 20\ \text{m s}^{-1} \[0.5em]0=u2−2×10×20u2=400⇒u=20 m s−1Hence, initial velocity of the ball = 20 m s-1(b) As we know, from the equation of motion,v2 = u2 + 2gsWhen the ball starts falling after reaching the maximum height, it's velocity (u) = 0s = 20 mg = 10 m s-2Substituting the values in the formula, we get,v2=02+(2×10×20)v2=400⇒v=20 m s−1\text v^2 = 0^2 + (2 \times 10 \times 20) \[0.5em] \text v^2 = 400 \[0.5em] \Rightarrow \text v = 20\ \text{m s}^{-1} \[0.5em]v2=02+(2×10×20)v2=400⇒v=20 m s−1Hence, final velocity of the ball on reaching the ground = 20 m s-1(c) As we know,total time of of journey of ball (t) = 2ug\dfrac{2\text u}{\text g}g2u​ andu = 20 m s-1g = 10 m s-2Substituting the values in the formula, we get,t=2×2010⇒t=4 s\text t = \dfrac{2 \times 20}{10} \[0.5em] \Rightarrow \text t = 4\ \text s \[0.5em]t=102×20​⇒t=4 sHence, total time for which the ball stays in air = 4 s

Question 15

Question 15(a) How long will a stone take to fall to the ground from the top of a building 80 m high and (b) what will be the velocity of the stone on reaching the ground ? (Take g = 10 m s-2 )

Accepted Answer

(i) From the equation of motion,h = ut + 12\dfrac{1}{2}21​ gt2Given,g = 10 m s-2h = 80 mu = 0Substituting the values in the formula, we get,80=(0×t)+(12×10×t2)80=0+(5×t2)80=5×t2t2=805t2=16t=4 s80 = (0 \times \text t) + (\dfrac{1}{2} \times 10 \times \text t^2) \[0.5em] 80 = 0 + (5 \times \text t^2) \[0.5em] 80 = 5 \times \text t^2 \[0.5em] \text t^2 = \dfrac{80}{5} \[0.5em] \text t^2 = 16 \[0.5em] \text t = 4\ \text s \[0.5em]80=(0×t)+(21​×10×t2)80=0+(5×t2)80=5×t2t2=580​t2=16t=4 sHence, time taken = 4 s(b) From the equation of motion,v2 = u2 + 2ghu = 0g = 10 m s-2h = 80 mSubstituting the values in the formula, we get,v2=02+(2×10×80)v2=1600⇒v=40 m s−1\text v^2 = 0^2 + (2 \times 10 \times 80) \[0.5em] \text v^2 = 1600 \[0.5em] \Rightarrow \text v = 40 \text{ m s} ^{-1}\[0.5em]v2=02+(2×10×80)v2=1600⇒v=40 m s−1Hence, final velocity of the stone on reaching the ground = 40 m s-1

Question 16

Question 16A body falls from the top of a building and reaches the ground 2.5 s later. How high is the building ? (Take g = 9.8 m s-2)

Accepted Answer

From the equation of motion,h = ut + 12\dfrac{1}{2}21​ gt2Given,g = 9.8 m s-2t = 2.5 su = 0Substituting the values in the formula, we get,h=(0×2.5)+(12×9.8×2.52)h=0+(4.9×2.52)h=4.9×(2.5)2h=30.6 m\text h = (0 \times 2.5) + (\dfrac{1}{2} \times 9.8 \times 2.5^2) \[0.5em] \text h = 0 + (4.9 \times 2.5^2) \[0.5em] \text h = 4.9 \times (2.5)^2 \[0.5em] \text h = 30.6\ \text m \[0.5em]h=(0×2.5)+(21​×9.8×2.52)h=0+(4.9×2.52)h=4.9×(2.5)2h=30.6 mHence, the height of the building is 30.6 m

Question 17

Question 17A ball is thrown vertically upwards with an initial velocity of 49 m s-1. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g = 9.8 m s-2)

Accepted Answer

(i) From the equation of motion,v2 = u2 - 2gh (a = -g , as the movement is against gravity)u = 49 m s-1v = 0g = 9.8 m s-2Substituting the values in the formula, we get,02=492−2×9.8×h2401=19.6×h⇒h=240119.6⇒h=122.5 m0^2 = 49^2 - 2 \times 9.8 \times \text h \[0.5em] 2401 = 19.6 \times \text h \[0.5em] \Rightarrow \text h = \dfrac{2401}{19.6} \[0.5em] \Rightarrow \text h = 122.5\ \text m\[0.5em]02=492−2×9.8×h2401=19.6×h⇒h=19.62401​⇒h=122.5 mHence, maximum height attained = 122.5 m(ii) As we know, total time of journey is,t = 2ug\dfrac{2\text u}{\text g}g2u​Substituting the values in the formula, we get,t=2×499.8t=989.8t=10 s\text t = \dfrac{2 \times 49}{9.8} \[0.5em] \text t = \dfrac{98}{9.8} \[0.5em] \text t = 10\ \text s \[0.5em]t=9.82×49​t=9.898​t=10 sHence, the time taken by the ball before it reaches the ground again = 10 s

Question 18

Question 18A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10 m s-2, calculate the height of the tower.

Accepted Answer

From the equation of motion,h = ut + 12\dfrac{1}{2}21​ gt2Given,g = 10 m s-2t = 4 su = 0Substituting the values in the formula, we get,h=(0×4)+(12×10×42)h=0+(5×16)h=5×16h=80 m\text h = (0 \times 4) + (\dfrac{1}{2} \times 10 \times 4^2) \[0.5em] \text h = 0 + (5 \times 16) \[0.5em] \text h = 5 \times 16 \[0.5em] \text h = 80\ \text m \[0.5em]h=(0×4)+(21​×10×42)h=0+(5×16)h=5×16h=80 mHence, the height of the tower is 80 m

Question 19

Question 19A pebble is dropped freely in a well from it's top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1, find: (i) the depth of water surface, and (ii) the time when echo is heard after the pebble is dropped.

Accepted Answer

(i) From the equation of motion,h = ut + 12\dfrac{1}{2}21​ gt2Given,g = 10 m s-2t = 20 su = 0Substituting the values in the formula, we get,h=(0×20)+(12×10×202)h=0+(5×400)h=5×400h=2000 m\text h = (0 \times 20) + (\dfrac{1}{2} \times 10 \times 20^2) \[0.5em] \text h = 0 + (5 \times 400) \[0.5em] \text h = 5 \times 400 \[0.5em] \text h = 2000\ \text m \[0.5em]h=(0×20)+(21​×10×202)h=0+(5×400)h=5×400h=2000 mHence, the height of the well is 2000 m(ii) As we know,time (t) = distancespeed\dfrac{\text{distance}}{\text{speed}}speeddistance​ort = depth of wellspeed of sound\dfrac{\text {depth of well}}{\text{speed of sound}}speed of sounddepth of well​Substituting the values in the formula, we get,t=2000330⇒t=6.1 s\text t = \dfrac{2000}{330} \[0.5em] \Rightarrow \text t = 6.1\ \text s \[0.5em]t=3302000​⇒t=6.1 sAs the pebble reaches the ground after 20 s, hence echo will be heard after (20 + 6.1) s = 26.1 s

Question 20

Question 20A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s-1. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9.8 m s-2

Accepted Answer

(i) As we know,v = u - gt (a = -g as the movement is against gravity)Given,u = 19.6 m s-1v = 0 (velocity on reaching the maximum height)g = 9.8 m s-2t = 5 sSubstituting the values in the formula, we get the time taken to reach the maximum height.0=19.6−9.8×t19.6=9.8×t⇒t=19.69.8⇒t=2 s0 = 19.6 - 9.8 \times \text t \[0.5em] 19.6 = 9.8 \times \text t \[0.5em] \Rightarrow \text t = \dfrac{19.6}{9.8} \[0.5em] \Rightarrow \text t = 2\ \text s \[0.5em]0=19.6−9.8×t19.6=9.8×t⇒t=9.819.6​⇒t=2 sHence, the time taken to reach the maximum height is 2 s and from maximum height back to the top of the tower = 2s.Therefore, time taken from the top of the tower to the ground = 5 - (2+2) = 1 s.So, with the help of the formula:h = ut + 12\dfrac{1}{2}21​ gt2we get,h=(19.6×1)+(12×9.8×12)h=19.6+4.9h=24.5 m\text h = (19.6 \times 1) + (\dfrac{1}{2} \times 9.8\times 1^2) \[0.5em] \text h = 19.6 + 4.9 \[0.5em] \text h = 24.5\ \text m \[0.5em]h=(19.6×1)+(21​×9.8×12)h=19.6+4.9h=24.5 mHence, height of the tower = 24.5 m(ii) Initial velocity (u) = 0 (on falling from maximum height)Time taken in falling from maximum height to ground = Total Time - Time taken to reach maximum height= (5 - 2) s= 3 sAs we know, v = u + gtSubstituting the values in the formula, we get,v=0+(9.8×3)⇒v=29.4 m s−1\text v = 0 + (9.8 \times 3) \[0.5em] \Rightarrow \text v = 29.4 \text { m s} ^{-1} \[0.5em]v=0+(9.8×3)⇒v=29.4 m s−1Hence, the velocity of ball on reaching the ground = 29.4 m s -1

Laws of Motion

Exercise 3C Multiple Choice Type

Exercise 3C Numericals

Exercise 3C Very Short Answer Type

Exercise 3E Assertion Reason Type

Exercise 3E Numericals