A 0.09kg ball is projected from a height of 1.46m with ahorizontal velocity of 1

Question

A 0.09kg ball is projected from a height of 1.46m with ahorizontal velocity of 1.8m/s and bounces from a 0.42kg smoothplate supported by springs. Knowing the rebound height is 0.55mdetermine
a) velocity of the plate immediately after impact b) the energy lost due to the impact A 0.09kg ball is projected from a height of 1.46m with ahorizontal velocity of 1.8m/s and bounces from a 0.42kg smoothplate supported by springs. Knowing the rebound height is 0.55mdetermine
a) velocity of the plate immediately after impact b) the energy lost due to the impact

Explanation / Answer

a) find velocity of the plate immediately after impact V:
vertical velocity of the ball before impact u:
u2 = 2gH, so u = (2gH) = 5.35 m/s (downward)
vertical velocity of the ball after impact u':
u'2 = 2gh, so u' = (2gh) = 3.28 m/s (upward)
during impact: mu = m(-u') + MV
V = m(u + u')/M = 1.85 m/s
b) find the energy lost due to the impact
mu2/2 - (mu'2/2 + MV2/2)
= mgH - mgh - MV2/2
= 0.0852 J

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