A50 kg person stands on a 25 kg sled. the sled and person are initially moving together at 6.0 m/s in the +x direction on a frictionless, horizontal surface. if the person jumps off the back of the sled and now slides at 2.0 m/s in the +x direction, how fast is the sled moving?

The sled is moving at 14 m/s

Explanation:

Using the law of conservation of momentum, we know that:

Total Momentum of system before collision = Total Momentum of system after collision

(m1)(u1) + (m2)(u2) = (m1)(v1) + (m2)(v2)

Here,

m1 = mass of the person = 50 kg

m2 = mass of sled = 25 kg

u1 = initial speed of the person = 6 m/s

u2 = initial speed of sled = 6 m/s

v1 = final speed of the person = 2 m/s

v2 = final speed of sled = ?

Therefore,

(50 kg)(6 m/s) + (25 kg)(6 m/s) = (50 kg)(2 m/s) + (25 kg)(v2)

(450 kg.m/s - 100 kg.m/s)/25 kg = v2

v2 = 14 m/s