A 65 kg ice skater traveling at 6.0ms^-1 runs head-on into an 85kg skater traveling forward at 4.5ms^-1. At what speed and in what direction do the skaters travel if they move together after the collision?

Question

A 65 kg ice skater traveling at 6.0ms^-1 runs head-on into an 85kg skater traveling forward at 4.5ms^-1.  At what speed and in what direction do the skaters travel if they move together after the collision?

anonymous · Answer

Since the skaters stick together, this is a perfectly inelastic collision. We can use the fact that momentum is conserved here. Momentum is defined as, $p = mv$. Let subscript a be the 65 kg skater and subscript b be the 85 kg skater. $$p_a - p_b = (m_a + m_b) v$$ This expression assumes the 65 kg skater has the positive velocity in the reference frame. Therefore, if the answer for the velocity after the collision is negative, the skaters will be travelling in the same direction as the 85 kg skater was initially, and vise-versa.

anonymous · Answer

I did (65+85)*4.5= 675

anonymous · Answer

We are solving for $v$. $$m_a * v_a - m_b*v_b = (m_a + m_b) v$$

anonymous · Answer

(65*6)-(85*4.5)=(65+85)v
7.5= 150v
v=.05

anonymous · Answer

Looks right.