A 2.71kg block traveling right at 3.46m/s collides with a 5.80kgblock traveling

Question

A 2.71kg block traveling right at 3.46m/s collides with a 5.80kgblock traveling left at 1.93m/s.
After the collision the 2.71kg block bounces back and is travelingat 4.01m/s.
What is the speed and direction of the 5.80kg block after thecollision?
Was energy lost or gained in the collision?
What type of collision was it? (elastic, inelastic, etc.)

Explanation / Answer

According to conservation of momentum we have            m1 u1 + m2 u2 = m1 v1 + m2 v2            2.7 *3.46 - 5.80 * 1.93 = -2.7 * 4.01 + 5.80 * v2 ==> v2 = 1.547 m/s     movingtowards right change in Energy = KEf - KEi                             = 0.5 [ m1 v12 + m2v22 - m1u12 -m2u22 ]                            = 0.5 [ 43.41 + 13.88 - 32.32 - 21.60 ]                             = 1.682 J Energy gained. Collision is elastic.

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