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A 0.5 kg ball moving with speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is
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- 24 N
- 12 N
- 96 N
- 48 N
Correct Option: A
Resolving the velocities in vertical and horizontal directions, resolved parts of first velocity 
v cosθ perpendicular to the wall and v sinθ parallel to the wall. In the second case, they are –v sin θ & v cos θ respectively. Here, –ve sign is because direction is opposite to the earlier ones. So we see a net change in velocity perpendicular to way
= v sin θ – (–v sin θ) = 2v sin θ
This change has occured in 0.25 sec, so, rate of change of velocity
| = | ||
| 0.25 |
| = | ⇒ | = 48 | ||
| 0.25 | 2 × 0.25 |
Thus, acceleration a = 48 m/sec²
Force applied = m . a = 0.5 × 48 = 24 N
