Question

1. A ball of mass 0.1kg is thrown against a wall. It strikes the wall normally with a velocity of 30m/s and rebounds with a velocity of 20m/s. calculate the impulse of the force exerted by the wall on the ball.

Answer 1

impulse=m(v-u)

=0.1(-20-30)

=0.1 x -50 =-5

Answer 2

Answer:

Impulse, J = -5 kg m/s.

Explanation:

It is given that,

Mass of the ball, m = 0.1 kg

It strikes the wall normally with a velocity of v₁ = 30 m/s and rebounds with a velocity of v₂ = -20 m/s.

We know that impulse is equal to the change in momentum.

So, $J=m\Delta V$

$J=0.1\times (v_2-v_1)$

$J=0.1\times (-20-30)$

$J=-5\ kg\ m/s$

So, the impulse of the force exerted by the wall on the ball is 5 kg m/s.