Question

A moving ball with a momentum of 25 kg m / s collides head-on with a wall.

It rebounds from the wall with the same speed but in the opposite direction. The time of the collision is 50 ms.
What is the average force exerted on the wall by the ball during the collision?
A 0.50N B 1.00N C 500N D 1000N
Show your working.

Answer 1

d 1000N

Step-by-step explanation:

change in momentum = 25 -(-25) = 50

by impuls momentum theorm

F.Δt = p2 - p1

f = p2 - p1 / Δt

f = 50/50*10^-3

so f = 1000N

Answer 2

Answer: D. 1000N

Step-by-step explanation:

p1 = 25 kgm/s, p2 = -25 kgm/s, t = 50 ms = 50 * 10^(-3) s = 0.05 s

change in momentum = initial momentum – final momentum

Mathematically represented as:

Δp = p1 - p2 = 25 - (-25) = 50 kgm/s

Δp = 50 kgm/s

change in momentum = Force * time interval

Mathematically represented thus:

Δp = F * Δt

Making F the subject of the equation, we have

F = Δp/Δt

F = 50/0.05

F = 1000 N