A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.460 m/s. The total mass of the sled, man, and rock is 95.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 15.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving).

0.41 m/s.

Explanation:

Mt = 95 kg

Mr = 0.31 kg

Vr = 15.5m/s

Ut = 0.46 m/s

Mass of the man and sled = (95 - 0.31) kg

= 94.69 kg

Using conservation of momentum equation,

Momentum before the throw = momentum after the throw

95 × 0.46 = 0.31 × 15.5 + 94.69 × V2

43.7 = 4.805 + 94.69 V2

V2 = 0.41 m/s.

Its velocity is (100m)/(10s)= 10 m/s right.

Explanation:

Since the motion is in a straight line, the displacement is equal to the distance. That makes the magnitude of the velocity equal to the speed.