Question

3.) A mass is moving at 9 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 0.23 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 hese two we got these put velocity in these formula with direction

- 6 = m * 9 +4*-v +4 *( -v-9) / ( m+4 ) ............1

13 = m * 9 +4*-v +m*(9--v) / ( m+4) ............... 2

from these two we got

m = 6.133 kg and v = -10 m/s

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 = 6.133 *( -6-9) /0.23 =-399.978261

so magnitude of force = **399.978261** **N
answer**

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

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