Question

a body of mass 2 kg moving with a velocity of 100m/s hits a wall and rebounds with the same speed if the contact time is 1 / 50 second the force applied on the wall is

Answer 1

The mass of the body is given as [m] = 2 kg.

The velocity of the body [v[ = 100 m/s

The initial momentum of the ball  $p_{i} =\ m\times v$

                                                               = 2 kg × 100 m/s

                                                               = 200 kg m/s

The final momentum of the ball $p_{f} =\ m\times v$

                                                           = 2 kg × (-100 m/s)

                                                           = -200 kg m/s.

Here, the velocity is taken negative as it is opposite to the initial velocity due to the bouncing of the body by wall.

The change in momentum will be -

                                          dp = $p_{f} -p_{i}$

                                               = -200 kg m/s - 200 kg m/s

                                               = -400 kg m/s

The contact time of the ball [dt] = $\frac{1}{50}\ s$

From Newton's second law , we know that -

                                      $F\ =\ \frac{dp}{dt}$

                                                 = $\frac{-400\ kg\ m/s}{0.02\ s}$

                                                 = - 20,000 N.

This is the force applied on body.

From Newton's third law, we know that action and reaction are equal and opposite.

Hence, the force applied on the wall is 20, 000 N.

Answer 2

mass =2kg

velocity = 100m/a

p(momentum)=m(mass)×v(velocity)

velocity=v+u=100+100(after rebounds)=200

p=2×200=400

IF TIME GIVEN THAN (P=MV÷T )

P=400×50/1=20000N=2×10⁴N is answer