CBSE Class 9 Science Question 18 of 25

Force and Laws of Motion — Question 14

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Question 14

An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Answer

Given,

Mass of the object (m1) = 1 kg

Mass of the block (m2) = 5 kg

Initial velocity of the object (u1) = 10 ms-1

Initial velocity of the block (u2) = 0

Mass of the resulting object = m1 + m2 = 6 kg

Velocity of the resulting object (v) = ?

Total momentum before the collision = m1u1 + m2u2 = (1 × 10) + 0 = 10 kgms-1

According to the law of conservation of momentum,

total momentum before the collision = total momentum after the collision.

∴ the total momentum post the collision is also 10 kgms-1

Hence, total momentum just before the impact = 10 kgms-1 and just after the impact = 10 kgms-1

Now, (m1 + m2) × v = 10 kgms-1

6 x v = 10

v = 106\dfrac{10}{6} = 53\dfrac{5}{3} ms-1

Hence, velocity of the combined object = 53\dfrac{5}{3} ms-1