A 0.4 kg ball traveling with the speed of 15 m/s strikes a rigid wall and rebounds elastically. If the ball is in contact with the wall for 0.045 s, what is (a) the momentum imparted to the wall and (b) the average force exerted on the wall?
(Ans. 12 kg m/s, 266.7N)


Given:

Mass of the ball = m = 0.4 kg
Initial Velocity of the ball before collision = vi = 15 m/s
Final Velocity of the ball after collision = vi = -15 m/s (being a elastic collision)
Time of contact of the ball with wall = t = 0.045 s

To Find:

(a) The momentum imparted to the wall = ΔP  = ?
(b) The average force exerted on the wall = F = ?

Solution:

(a) the momentum imparted to the wall:

According to the laws of conservation of momentum for an elastic collision in the questioned case can be written as

ΔP ball = -ΔP wall --------------(1)

According to the given data, we can find the Î”P ball, i.e  

ΔP = Pf - Pi = m vf - m vi = m ( vf - vi)

by putting values

ΔP = 0.4 kg ( -15 m/s - 15 m/s)

ΔP = 0.4 kg ( -30 m/s)

ΔP = -12 kg m/s

Now using equation (1)

-12 kg m/s = -ΔP wall

or

ΔP wall =  12 kg m/s

thus the momentum imparted to the wall is 12 kg m/s


(b) the average force exerted on the wall?

We know that a force in terms of momentum is defined as the rate of change in momentum. Mathematically it is written as

F = `\frac {ΔP}{t}` 

Now the average force exerted on the wall is

F wall = `\frac {ΔP wall}{t}` 

F wall = `\frac {12 kg m s^{-1}}{0.045 s}`

After simplifying we get

F wall = 266.666 N

F wall = 267 N

Thus the average force exerted on the wall is 12 kg m/s


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