Question

A mass is moving at 6 m/s in the +x direction and it collides in a perfectly elastic collision with a mass of 5 kg moving in the -x direction. The collision takes places in 0.25 seconds and after the collision the mass that was moving in the +x direction is moving in the -x direction at 6 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 we apply momentum and energy conservation to get these

m1*v1 +m2*v2 = m1*v1f + m2*v2f   

and 1/2*m1*v1^2 +1/2*m2*v2^2 = 0.5*m1*v1f^2 +0.5*m2*v2f^2

from these two we got these put velocity in these formula with direction

now we need to plug the values in these equations to get the mass of the first particle

-6 = ( m *6 +5*-v +5*(-v-6) /( m+5) .........1

13 = ( m*6 +5*-v +m*(6--v) /(m+5) ........2

from 1 and 2 we got

m = 10.833 and v = -13 m/s (mass of 5 kg velocity)

What is the magnitude of the average force, in Newtons, on the first mass

F =dp/dt = change in momentum /time   

= m * ( V1f -v1i ) /t = 10.833 *( -6-6) /0.25 = -519.98400 N

magnitude = 519.984 N

let me know in a comment if there is any problem or doubts