A 260 g ball is dropped from a height of 2.4 m , bounces on a hard floor, and re

Question

A 260 g ball is dropped from a height of 2.4 m , bounces on a hard floor, and rebounds to a height of 1.9 m . (Figure 1)  shows the magntiude of the normal force exerted on the ball while it is contact with the floor. .

Part A: With what velocity does the ball hit the ground? Express your answer to three significant figures and include the appropriate units. Use the usual convention that up is positive. Vy=_____

Part B: With what velocity does the ball leave the ground after the bounce? Express your answer to three significant figures and include the appropriate units. Use the usual convention that up is positive. Vy=____

Part C: What total impulse is imparted to the ball during the bounce? Express your answer to three significant figures and include the appropriate units. J=_____ (kg*m/s)

Part D: If the ball is on contact with the floor for a time t = 4 ms , what maximum force, Fmax, does the floor exert on the ball?  Hint: Given the small amount of time and the small mass of the ball, the impulse imparted the ball's weight is negligible.Express your answer using three significant figures and include the appropriate units. Fmax=_____

Explanation / Answer

part A

along vertical

initial velocity voy = 0

acceleration ay = -g = -9.8 m/s^2

displacement y = -2.4 m

finalvelocity vy = /

vy^2 - voy^2 = 2*ay*y

vy^2 - 0 = 2*9.8*2.4

vy = -6.86 m/s

part B

for upward motion

along vertical

initial velocity = v2y

acceleration ay = -g = -9.8 m/s^2

displacement y2 = 1.9 m

finalvelocity v3y = 0

v3y^2 - v2y^2 = 2*ay*y

0 - v2y^2 = -2*9.8*1.9

v2y = 6.1 m/s

part (C)

impulse = change in momentum

J = m*(v2y-vy)

J = 0.26*(6.1 - (-6.86)) = 3.37 kg m/s

======================

part D

maximum force Fmax = J/t = 842 N <<<<-----------ANSWER

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