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QAIPMT-2006-Ph-2454: NEET 2006 Physics - Class 11 AIPMT-2006-Ph-2454 Solution: A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall …
A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of with the wall. It is reflected with the same speed 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:
Multiple Choice Options
- A 96 N
- B 48 N
- C 24 N
- D 12 N
Step-by-Step Verified Solution
Resolving the velocities in vertical and
horizontal directions, resolved parts of first
velocity
v cos _ perpendicular to the wall and v sinq
parallel to the wall. In the second case, they are
�v sin _ & v cos _ 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 _ � (�v sin _ ) = 2v sin _
This change has occured in 0.25 sec, so, rate of
change of velocity = 2 v sin _ _ 0.25
= 2 _ 12 _ sin _ 30 _ 0.25 _ 24 _ 1 2 _ 0.25 = 48
Thus, acceleration a = 48 m/sec 2
Force applied = m _ a = 0.5 _ 48 = 24 N