Question

A ball of mass $1kg$ is dropped from $20m$ height on ground and it rebounds to height $5m$ . Find magnitude of change in momentum during its collision with the ground. (Take $g = 10m{s^{ - 2}}$ )

Answer 1

Hint: Calculate the velocity before the collision and after the collision. Use the equation for change in momentum and take the magnitude. Remember to use proper signs for the velocity of approach and the velocity of separation.

Complete step by step answer:
We take the downward direction as negative and the upper direction as positive.
The initial velocity or the velocity of approach is given by
\[
u = - \sqrt {2gh} \\
\Rightarrow u = - \sqrt {2 \times 10 \times 20} \\
\Rightarrow u = - 20m{s^{ - 1}} \\
\]
We take this negative because it is going down
The final velocity after the collision is given by
$
v = \sqrt {2gh} \\
\Rightarrow v = \sqrt {2 \times 10 \times 5} \\
\Rightarrow v = 10m{s^{ - 1}} \\ $
We take this positive because it is going up
Thus the change in momentum is given by
$
\Delta p = m(v - u) \\
\Rightarrow\Delta p = 1(10 - ( - 20)) \\
\therefore\Delta p = 30kgm{s^{ - 1}} \\$
Hence the change in momentum is given by $\Delta p = 30kgm{s^{ - 1}}$.

Additional Information:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. It is a vector quantity. Momentum is the quantity of motion contained in an object. It is the product of mass and velocity. It is a vector quantity. The change of momentum is known as impulse while the rate of change in momentum is called force.

Note:We use basic kinematics to get the value of the velocity of approach and velocity of separation. Take care of the signs of the velocity as both will be opposite signs since both are going in opposite directions.