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 2 kg moving in the -x direction. The collision takes places in 0.21 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 13 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 = 2 kg

speeds before collision

v1i = 10 m/s                                 v2i = 0
?

speeds after collision

v1f = -9 m/s                                        v2f = 13 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)

-9 = ((m1-2)*10 + 2*2*v2i)/(m1+2)............(3)

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

13 = ((2 - m1)*v2i + 2*m1*10)/(m1+2)..................(4)

solving above equations 3 and 4

m1 = 2.63 kg

v2i = -12 m/s

F1 = m1*(V1f - v1i)/t

F1 = 2.63*(-9-10)/0.21

F1 = 238 N