Question

7. A ball of mass 20 gram hits a wall at an angle of
45° with a velocity of 15 m/s. If the ball rebounds
at 90° to the direction of incidence, calculate the
impulse received by the ball.
​

Answer 1

Answer:

Answer

m=20g=0.02kg,

v=15m/s

Impulse is given by momentum change =m.Δv

=m.(vf-vi )sinθ

=0.02(−15−15)sin45

=0.42kg.m/s

Answer 2

Answer:

The impulse received by the ball is 0.8484Ns.

Explanation:

m = 20g = 0.02kg theta = 45° v = 15m/s

Impulse(I) = 2mvsin(theta)

{ In this case where velocity of the ball remains same the impulse received by the ball is always 2mvsin(theta)}

= 2 × 0.02 × 15 × sin45°

= 2 × 0.02 × 15 × 1/root2

= root2 × 0.02 × 15

= 0.3root2

= 0.3 × 1.414

= 0.8484