A skateboarder coasting at 5.0 m>s decides to coast into a friend going 4.0 m>s

Question

A skateboarder coasting at 5.0 m>s decides to coast into a
friend going 4.0 m>s on rollerblades in the same direction.
Unfortunately, the rollerblader stops right before the collision,
too quickly for the skateboarder to react. The rollerblader
turns around as he stops and pushes his arms out to
ward off the collision. The rollerblader and his gear have a
combined inertia of 70 kg, and the skateboarder and his gear
have a combined inertia of 79 kg.

(d) If the rollerblader had not turned around and
stopped, so that the skateboarder was able to grab him and
keep coasting, what amount of kinetic energy would have
been lost in the collision?

answer is d) 2×10 J loss in KE to internal energy

could you explain for part d?

Explanation / Answer

part d) in that case , if rollerblader has not stopped ,

let the velocity after the collision will be v

Using conservation of momentum

70 * 4 + 79 * 5 = (70 + 70) * v

v = 4.53 m/s

loss in kinetic energy = initial kinetic energy - final kinetic energy

loss in kinetic energy = 0.50 * 79 * 5^2 + 0.50 * 70 * 4^2 - 0.50 * (70 + 79) * 4.53^2

loss in kinetic energy = 19 J

rouding the value to one significant figure

loss in kinetic energy = 20 J

the loss in kinetic energy is 20 J

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