Question

A mass is moving at 10 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 0.22 seconds and after the collision the mass that was moving in the +x direction is moving in the -x direction at 8 m/s and the mass that was moving in the -x direction is moving in the +x direction at 14 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

ELASTIC COLLISION

m1 = ?                                     m2 = 4 kg

speeds before collision

v1i = 10 m/s m/s                                 v2i = -v1

speeds after collision

v1f = - 8 m/s                                        v2f = + 14 m/s

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

---------------------

-8 = (( m1 - 4)*10 - (2*4*v2))/(m1+4)

-8*m1 - 32 = 10*m1 - 40 - 8*v2

18*m1 - 8 - 8*v2 = 0

18*m1 = 8*v2 + 8..............(1)

14 = ( -( 4 - m1)*v2 + (2*m1*10))/(m1+4)

14*m1 + 56 = - 4*v2 + v2*m1 + 20*m1

m1*(6 + v2) - 4*v2 - 56 = 0

m1*(6 + v2) = 4*v2 + 56 = 0 .................(2)

18/(6+v2) = (8v2 + 8)/(4*v2 + 56)

v2 = 12 m/s

m1 = 5.78 kg

=================

force F = m1*(v1f - v1i)/t

F = 5.78*(-8-10)/0.22 = -473 N

magnitude = 473 N