Question 5A bullet initially moving with a velocity 20 m s-1 strikes a target and comes to rest after penetrating a distance 10 cm in the target. Calculate the retardation caused by the target.

Question

Accepted Answer

According to equation of motion —v2 = u2 + 2aSGiven,u = 20 m s-1,v = 0,s = 10 cm = 0.1 mSubstituting the values in the formula above, we get,0=(20)2+(2×a×0.1)⇒0=400+0.2a⇒−400=0.2a⇒a=−4000.2⇒a=−40002⇒a=−2000 ms−20 = (20)^2 + (2\times \text a \times 0.1) \[0.5em] \Rightarrow 0 = 400 + 0.2 \text a \[0.5em] \Rightarrow -400 = 0.2\text a \[0.5em] \Rightarrow \text a = -\dfrac{400}{0.2} \[0.5em] \Rightarrow \text a = -\dfrac{4000}{2} \[0.5em] \Rightarrow \text a = - 2000\ \text {ms}^{-2}\[0.5em]0=(20)2+(2×a×0.1)⇒0=400+0.2a⇒−400=0.2a⇒a=−0.2400​⇒a=−24000​⇒a=−2000 ms−2Hence, Retardation is 2000 m s-2

Motion in One Dimension — Question 5

Question 5