Question

A cart with mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at .66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two cart center of mass?

Please explain the derivation of any substitutions. This is where I am having the most trouble.

Answer 1

masses m = 340 g

            M = ?

Initial speed of 1 st cart u = 1.2 m / s

initial speed of 2 nd cart U = 0

final speed of 1 st cart v = 0.66 m / s

for eleastic collision ,coeffcient of restitution e= 1

       e = ( V - v ) / ( u - U )

      1 = ( V - 0.66) / ( 1.2-0)

V-0.66 = 1.2

      V = 1.86 m / s

(a). from law of conservation of momentum ,

mu + MU = mv + MV

m( u -v) = M ( V - U )

               = MV

from this mass of 2nd cart M = [ m(u-v) ] / V

                                             = 98.7 g

(b). speed of cart2 after impact V = 1.86 m / s

(c). speed of the two cart center of mass V ' = [mv + MV ] / ( m + M )

                V ' = 0.93 m / s

Answer 2

Answer 3

Thats just a conservation of momentum problem. the mass of the mass of the cart times the velocity of the first cart equals the mass times the velocity plus the mass times the velocity of both the carts combined and if they don't hit head on it's the mass times the velocity devided by the cos of the angle for the first cart and mass times velocity times the cos of the angle for the second cart