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Force Ball Point
0.640 kg ball is dropped from rest at a point 1.05 m above the ground the ball bounces back ..? Help pls!?
0.640 kg ball is left falling from rest at a point 1.05 m above the ground. The ball bounces up to a height of 0,630 m. What are the magnitude and direction of the force pulse net applied to the ball during the collision with the ground?
Find the speed at which it hits the ground: v ^ 2 = u ^ 2 + 2as v ^ 2 = 2 = 0 + v = 4.54ms 2×9.8×1.05 ^ -1 Therefore, the momentum just before the collision = 0.64×4 .54 = 2.9kgms ^ -1 when it bounces at the speed of u, v ^ 2 ^ u = 2 + 2as 0 ^ u = 2 – 2×9.8x.63 v = 3.51ms ^ -1 Therefore, the final push on collision 2.25kgms 0.64×3.51 = ^ = -1 Therefore, the change in momentum = 2.9 – (-2.25) (momentum is a vector) = 5.15kgm ^ s-1 is the same impulse.
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