Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The total mass of the sled, man, and rock is 93.5 kg. The mass of the rock is 0.320 kg and the man can throw it with a speed of 16.5 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) in m/s

Determine the speed of the sled (in m/s) if the man throws the rock directly backward in m/s

Answer 1

here,

the mass of man and sled , m1 = 93.5 kg

mass of rock , m2 = 0.32 kg

initial speed , u = 0.48 m/s

when the rock is thrown with v2 = 16.5 m/s in forward direction

let the final speed of the sled be v1

using conservation of momentum

(m1 + m2) * u = m1 * v1 + m2 * v2

(93.5 + 0.32) * 0.48 = 93.5 * v1 + 0.32 * 16.5

solving for v1

v1 = 0.43 m/s

the final speed of sled is 0.43 m/s

when the rock is thrown with |v2| = 16.5 m/s in backward direction

v2 = - 16.5 m/s

let the final speed of the sled be v1

using conservation of momentum

(m1 + m2) * u = m1 * v1 + m2 * v2

(93.5 + 0.32) * 0.48 = 93.5 * v1 - 0.32 * 16.5

solving for v1

v1 = 0.51 m/s

the final speed of sled is 0.51 m/s