Question

A mass is moving at 8 m/s in the +x direction and it collides in a perfectly elastic collision with a mass of 4 kg moving in the -x direction. The collision takes places in 0.20 seconds and after the collision the mass that was moving in the +x direction is moving in the -x direction at 9 m/s and the mass that was moving in the -x direction is moving in the +x direction at 11 m/s. What is the magnitude of the average force, in Newtons, on the first mass which was originally moving in the +x direction before the collision?

Answer 1

given

u1 = 8 m/s

m2 = 4 kg

after the collision,

v1 = -9 m/s

v2 = 11 m/s

let m1 is the mass of the first body and u2 is the initial velocity of the second body.

we know,

v1 = ( (m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)

(-9) = ( (m1 - 4)*8 + 2*4*u2 )/(m1 + 4) ----(1)

we know,

v2 = ( (m2 - m1)*u2 + 2*m1*u1 )/(m1 + m2)

11 = ( (4 - m1)*u2 + 2*m1*8 )/(m1 + 4) ----(2)

on solving above two equationms we get

m1 = 5.4 kg

u2 = -12 m/s

magnitude of average force acting on first body, |F_avg| = m1|a|

= m1*(u1 - v1)/t

= 5.4*(8 - (-9))/0.2

= 459 N <<<<<<<<<<<---------------Answer