Question

A ball of mass $1\text{kg}$ strikes a heavy platform, elastically, moving upwards with a velocity of $5\text{m/s}$. The speed of the ball just before the collision is $10\text{m/s}$ downwards. Then the impulse imparted by the platform on the ball is (in $\text{N-s}$)

Answer 1

Given velocity of platform is $u=5\text{m/s}$ and initial velocity of ball is $v_1=-10\text{m/s}$,(Considering velocity of platform as positive)
let $v_2$ be final velocity of the ball after collision,
As platform is heavy w.r.t the ball, velocity of platform doesnot change.
As, collision is elastic $e=1=\frac{v_2-u}{u-v_1}$
$v_2=2u-v_1$
$v_2=2\times 5+10=20\text{m/s}$
Now Impulse on the ball is
$\overrightarrow{J}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i=1\times 20-(-1\times 10)=30\text{N-s}$