MCAT Physics Question — Ice Skaters
 by
 May 26, 2014
 MCAT Blog, MCAT Physics, MCAT Question of the Day
A motionless, inexperienced ice skater of 70kg faces forward with skates parallel. A more experienced skater of 80kg collides with the stationary skater at a speed of 2m/sec.Both skaters move off together with the same velocity. If ice is considered frictionless, what is the approximate speed of the two skaters after the collision?
a) 1.1m/sec
b) 1.4 m/sec
c) 2 m/sec
d) 4 m/sec
Explanation
When two masses collide and move with the same velocity after impact, conservation of momentum must be applied. m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{final}, where v_{1}and v_{2} are the initial velocities of masses 1 and 2. Here, skater 1 has mass of 70kg and initial velocity of 0m/sec. Skater 2 has a mass of 80kg and initial velocity of 2m/sec.
Applying conservation of momentum, v_{final}=m_{2}/( m_{1}+m_{2})*v_{2}. From looking at the numbers, v_{final} is approximately 1/2 of v_{2} because the skaters have roughly the same mass. Choice a is correct.
a) 1.1m/sec, correct.
b) 1.4 m/sec, incorrect, This assumes kinetic energy before and after is equal. Energy is not conserved in an inelastic collision.
c) 2 m/sec, incorrect, This assumes that speed stays constant after a collision.
d) 4 m/sec, incorrect, This is the initial velocity of the experienced skater squared.
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