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.

Asked by Vishal kumar | 1 year ago |  289

1 Answer

Solution :-

Given, mass of the object (m1) = 1 kg

Mass of the block (m2) = 5 kg

Initial velocity of the object (u1) = 10 m/s

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 kg) × (10 m/s) + 0 = 10 kg.m.s-1

As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s-1

Now, (m1 + m2) × v = 10 kg.m.s-1

Therefore, v = \( \frac{10\,kg.m.s^{-1}}{6\,kg}=1.66ms^{-1}\)

The resulting object moves with a velocity of 1.66 meters per second.

Answered by Shivani Kumari | 1 year ago

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