Question

A ball having mass 500g hits wall with a 10m/s velocity. Wall applies 4000 N force to the ball and it turns back with 8m/s velocity. Find the time of ball-wall contact.​

Answer 1

Initial momentum=mv= 1/2*10=5kgm/s

Final= m(-v)= -mv= -1/2(8)= -4kgm/s

Δmomentum=final-initial = -4 - 5 = -9kgm/s

F= Δmomentum/Δtime

Δtime=Δmomentum/F

        =9/4000 seconds.

Hence t= 9/4000= 2.25*10⁻³ seconds

Answer 2

Answer:

The time of ball-wall contact is 2.5×10⁻⁴seconds.

Explanation:

We will use the first equation of motion to solve this question,

$v=u+at$     (1)

Where,

v=final velocity of the ball

u=initial velocity of the ball

a=acceleration of the ball

t=time taken for the changes

From the question we have,

u=10m/s

v=8m/s

Mass of the ball(m)=500g=0.5kg

Force applied by the wall(F)=4000N

And we know that the force is given as,

$F=ma$    (2)

$a=\frac{4000}{0.5} =8000m/s^2$   (3)

By substituting the required values in equation (1) we get;

$8=10+8000\times t$

$t=\frac{8-10}{8000}=\frac{-2}{8000}=-0.00025sec$  

(-ve sign only indicates that the direction of the velocity of the ball has changed)

$t=2.5\times 10^-^4sec$      (4)

Hence, the time of ball-wall contact is 2.5×10⁻⁴seconds.