266_PartUniversity Physics Solution

# 266_PartUniversity Physics Solution - Momentum Impulse and...

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Momentum, Impulse, and Collisions 8-25 EVALUATE: The tension before the collision is the weight of the ball, 78.4 N. Just after the collision, when the ball has started to move, the tension is greater than this. a rad mg y x T Figure 8.81 8.82. IDENTIFY: The impulse applied to the ball equals its change in momentum. The height of the ball and its speed are related by conservation of energy. SET UP: Let + y be upward. EXECUTE: Applying conservation of energy to the motion of the ball from its height h to the floor gives 2 1 1 2 mv mgh = , where 1 v is its speed just before it hits the floor. Just before it hits, it is traveling downward, so the velocity of the ball just before it hits the floor is 1 2 y vg h =− . Applying conservation of energy to the motion of the ball from just after it bounces off the floor with speed 2 v to its maximum height of 0.90 h gives 2 1 2 2 (0.90 ) mv mg h = . It is moving upward, so 2 2 (0.90 ) y h =+ . The impulse applied to the ball is 21 () yyy y y Jp pm vv = −= − = 2 (0.90 ) 2 2.76 mg hmg h m g h += . The floor exerts an upward impulse of 2.76 h to the ball. EVALUATE: The impulse increases when m increases and when h increases. The ball does not return to its initial height because some mechanical energy is dissipated during the collision with the floor. 8.83. IDENTIFY: Apply conservation of momentum to the collision between the bullet and the block and apply conservation of energy to the motion of the block after the collision. (a) SET UP: Collision between the bullet and the block: Let object A be the bullet and object B be the block. Apply momentum conservation to find the speed 2 B v of the block just after the collision. Figure 8.83a EXECUTE: x P is conserved so 11 2 2 AAx BBx mv +=+ . 122 AA . ( ) 3 12 2 4.00 10 kg 400 m/s 120 m/s 1.40 m/s 0.800 kg A Bx B mv v v m −× == = . SET UP: Motion of the block after the collision. Let point 1 in the motion be just after the collision, where the block has the speed 1.40 m/s calculated above, and let point 2 be where the block has come to rest. Figure 8.83b 1 1 other 2 2 K UW KU ++ = + . EXECUTE: Work is done on the block by friction, so other . f WW = other k k cos , fk WWf s f s m g s φμ = = where 0.450 m s = 0, 0 UU 2 1 11 2 2 ,0 Km v K (block has come to rest) Thus 2 1 1k 2 0. mv mgs μ 2 2 1 k 2 1.40 m/s 0.222 2 2 9.80 m/s 0.450 m v gs = .

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8-26 Chapter 8 (b) For the bullet, ( ) () 2 23 11 22 4.00 10 kg 400 m/s 320 J Km v == × = . ( ) 2 4.00 10 kg 120 m/s 28.8 J v × = . 21 28.8 J 320 J 291 J KK K Δ= − = = . The kinetic energy of the bullet decreases by 291 J. (c) Immediately after the collision the speed of the block is 1.40 m/s so its kinetic energy is 2 2 0.800 kg 1.40 m/s 0.784 J. v = EVALUATE: The collision is highly inelastic. The bullet loses 291 J of kinetic energy but only 0.784 J is gained by the block. But momentum is conserved in the collision. All the momentum lost by the bullet is gained by the block. 8.84. IDENTIFY: Apply conservation of momentum to the collision and conservation of energy to the motion of the block after the collision.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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266_PartUniversity Physics Solution - Momentum Impulse and...

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