Question

Two carts are on a flat surface of negligible friction. Cart A has a mass of 328 [grams] and cart B has a mass of 315 [grams]. Then an object of unknown mass is placed in one cart A and the two carts experience a completely inelastic collision. At the instant before collision Cart B moves .3m in .3 seconds and Cart A with it unknown mass placed on it, is motionless. After the collision Cart B + Cart A move .2m in approximately 1.23 seconds. calculate the experimental value for the item of the unknown mass.

Answer 1

There is no horizontal force on the carts and hence the momentum of the system is conserved.

m1u1 + m2u2 + mu = m1v1 + mv2 + mv3  -------------1

m1 is mass of cart A and u1 is its speed before collision (u1 = 0)

m2 is mass of cart B and u2 is its speed before collision

(u2 = distance / time = 0.3m / 0.3s => u2 = 1m/s)

m is unknown mass in cart A and u is its speed (which equals to u1 = 0, as it is in cart A)

Since the collision is pure inelastic then all the objects (cart A + cart B + unknown mass) will have same mass. v1 = v2 = v3 = V

V = distance / time = 0.2m / 1.23s => V = 0.163 m/s

From equation 1,

0.328 x 0 + 0.315 x 3 + m x 0 = (0.328 + 0.315 + m) x 0.163

0.945 = (0.643 + m) x 0.163

5.8 = 0.643 + m

m = 9.02 kg = 9020 g