Question

Mass m = 0.1 kg moves to the right with speed v = 0.48 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction = 0.88 of its original kinetic energy. If the masses remain in contact for 0.01 secs while colliding, what is the magnitude of the average force in N between the masses during the collision? Hints: All motion is in 1D. Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. VR must be greater than VL since the masses can't pass through each other!

Answer 1

let
m1 = m2 = m = 0.1 kg
u1 = 0.48 m/s

KEi = (1/2)*m1*u1^2

let v1 and v2 are velocities of m1 and m2 after the collision.

KEf = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= (1/2)*m*(v1 + v2)^2 (since m1 = m2 = m)

Apply conservation of momentum

m1*u1 = m1*v1 + m2*v2

u1 = v1 + v2

0.48 = v1 + v2 ----(1)

given

KEf = 0.88*KEi

(1/2)*m*(v1^2 + v2^2) = 0.88*(1/2)*m1*u1^2

(v1^2 + v2^2) = 0.88*0.48^2

(v1^2 + v2^2) = 0.202752 ------(2)

on solving equations 1 and 2

we get
v1 = 0.0308 m/s
v2 = 0.449 m/s

Let F is the magnitude of average force between the masses during the collision,

use Impulse = change in momentum

F*t = change in momentum of m2

F = change in momentum of m2/t

= m2*(v2 - u2)/t

= 0.1*(0.449 - 0)/0.01

= 4.49 N <<<<<<<<<-----------Answer