Question

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.490 m/s. The total mass of the sled, man, and rock is 95.0 kg. The mass of the rock is 0.260 kg and the man can throw it with a speed of 16.0 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).

_______m/s

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

_______m/s

Answer 1

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

using conservation of momentum, we have

p_before = p_after

m_total v_sled,i = m_sled v_sled + m_rock v_rock,f

m_total v_sled,i = (m_total - m_rock) v_sled + m_rock v_rock,f

(95 kg) (0.490 m/s) = [(95 kg) - (0.260 kg)] v_sled + (0.260 kg) (16 m/s)

(46.55 kg.m/s) = (94.74 kg) v_sled + (4.16 kg.m/s)

[(46.55 kg.m/s) - (4.16 kg.m/s)] = (94.74 kg) v_sled

v_sled = [(42.39 kg.m/s) / (94.74 kg)]

v_sled = 0.447 m/s

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

using conservation of momentum, we have

p_before = p_after

m_total v_sled,i = m_sled v_sled + m_rock v_rock,f

m_total v_sled,i = (m_total - m_rock) v_sled + m_rock v_rock,f

(95 kg) (0.490 m/s) = [(95 kg) - (0.260 kg)] v_sled + (0.260 kg) (-16 m/s)

(46.55 kg.m/s) = (94.74 kg) v_sled - (4.16 kg.m/s)

[(46.55 kg.m/s) + (4.16 kg.m/s)] = (94.74 kg) v_sled

v_sled = [(50.71 kg.m/s) / (94.74 kg)]

v_sled = 0.535 m/s