After a 0.200-kg rubber ball is dropped from a height of 1.80 m, it bounces off

Question

After a 0.200-kg rubber ball is dropped from a height of 1.80 m, it bounces off a concrete floor and rebounds to a height of 1.35 m. Determine the magnitude and direction of the impulse delivered to the ball by the floor, magnitude kg m/s direction Estimate the time the ball is in contact with the floor to be 0.03 seconds. Calculate the average force the floor exerts on the ball, magnitude N direction A railroad car of mass 2.55 Times 104 kg is moving with a speed of 4.10 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.05 m/s. What is the speed of the four cars after the collision? m/s How much mechanical energy is lost in the collision? A baseball approaches home plate at a speed of 48.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 50.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.80 ms. What is the average vector force the ball exerts on the bat during their interaction? F on bat = N

Explanation / Answer

1)a)change in momentum = 0.2 * ( 5.14 - 5.94 ) = -0.1599

b) F = - 5.33 N (upwards )

2) m * 4.1 + 3*m*2.05 = 4*m*v
v = 2.5625 m/s

b)

mechanical energy lost = m [(0.5*4*2.5625*2.5625 )-(0.5*4.1*4.1 )-(0.5*3*2.05*2.05)]=
= -40186.4 J

3)

U = 48 i
V = 50 j

F *1.8*10^-3 = 0.145 * 69.3

F =38506.052 N

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