What is the resulting speed of the combined man and sled?

Question

A man of mass 60 kg running with a constant
speed 5.0 m/s jumps onto a sled of mass 30 kg
which is moving with a speed 2.0 m/s on the
frictionless ice. What is the resulting speed of
the combined man and sled? Note that both the
man and the sled are initially moving in the
same direction.

 

The answer should be 4 m/s

 

Thank you for your solution!!

Answer 1

Conservation of momentum states that the initial momentum of a system must equal its final momentum. (p_i=p_f)

 

By definition, momentum is the mass of a moving object times its velocity. (p=m*v)

 

So, since p_i=p_f, and the man and sled are one coupled object in the final state, we can set (p_i-man + p_i-sled) = p_f

 

Substituting the appropriate (m*v) for each p in the equation, we get:

 

(m_i-man*v_i-man) + (m_i-sled*v_i-sled) = m_f*v_f

 

60*5.0 + 30*2.0 = (60+30)*V

300 + 60 = 90V

360 = 90V

4 = V

 

When we calculate the units, we have units of kg in every term, so we can divide through by kg and have no mass in the final solution. All the terms will have units of m/s. When we add units of the same type, we carry the units through. So our final solution is 4 m/s.