A0.7 kg lab cart moving to the right at 0.15 m/s collides with a 0.5 kg lab cart moving to the right at 0.10 m/s. after the collision, the 0.7 kg cart is moving to the right at 0.08 m/s. calculate the velocity of the 0.5 kg lab cart after the collision.

0.198 m/s

Explanation:

m1 = 0.7 kg, u1 = 0.15 m/s,

v1 = 0.08 m/s

m2 = 0.5 kg, u2 = 0.1 m/s

Let the speed of 0.5 kg is v2 after the collision.

By using the conservation of momentum

Momentum before collision = momentum after collision

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.7 x 0.15 + 0.5 x 0.1

= 0.7 × 0.08 + 0.5 × v2

0.099 = 0.5 v2

v2 = 0.198 m/s

answer: the answer is a

f=ma newton's second law 7x8=56n