Question

a 0.4kg ball travelling with the speed of 15m/s strikes a rigid wall and rebounds elastically. If the ball is in contact with the wall for 0.045s,what is (a) momentum imparted to the wall and (b) average force exerted on the wall? ​

Answer 1

Answer:

Explanation:

Mass = 0.4 kg

velocity = 15m/s

Momentum of the ball = mv = 6kgm/s

According to the law of conservation of momentum,

Momentum is conserved.

So the momentum imparted to the wall = 6kgm/s

Force = momentum/time

          = 6/0.045 = 6000/45 = 400/3

          = 133.3 N

Answer 2

Answer:

Data:

Mass of ball, m= 0.4kg

velocity of ball, v=15m/s

Δt= 0.045 s

To Find:  

a) Momentum imparted to the wall , P =?

b) Average Force, $F_{av}$= ?

Solution:

Momentum before collision, $P_{1}$= mv

                                                  = 0.4 ×15

                                                   = 6 kg m/s

As ball rebounds elastically,

Momentum after collision, $P_{2}$= - mv

                                                 = -0.4 ×15

                                                 =  -6 kg m/s

Change in momentum= ΔP

                                      = $P_{2} -P_{1}$

                                      =-6 - 6

                                      = -12 kg m/s

a) Momentum imparted to the wall, P = - ΔP

                                                               = - (-12)

                                                                = 12 kg m/s

b)    Δ$F_{av} =$ ΔP

                    Δt

                =$\frac{12}{0.045}$

               = 266.7 N