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Physics 192 Solutions to Mastering Physics Week 1 P9.3. Prepare: From Equations 9.5, 9.2, and 9.8, Newton’s second law can be profitably rewritten as p Favg t In fact, this is much closer to what Newton actually wrote than F ma. Solve: This allows us to find the force on the snowball. By Newton’s third law we know that the snowball exerts a force of equal magnitude on the wall. p mvf mvi m(vf vi ) (0.12 kg)(0 m/s 7.5 m/s) Favg 6.0 N t t t 0.15 s where the negative sign indicates that the force on the snowball is opposite its original momentum. So the force on the wall is also 6.0 N. Assess: This is not a large force, but the snowball has low mass, a moderate speed, and the collision time is fairly long. P9.7. Prepare: Please refer to Figure P9.7. Model the object as a particle and its interaction with the force as a collision. We will use Equations 9.1 and 9.9. Because p = mv, so v = p/m. Solve: (a) Using the equations ( px )f ( px )i Jx J x area under the force curve (vx )f 1.0 m/s (1.0 m/s) 1 (area under the force curve) 2.0 kg 1 (1.0 N s) 1.5 m/s 2.0 kg (b) Likewise, 1 1 (vx )f (1.0 m/s) (area under the force curve) (1.0 m/s) ( 1.0 N s) 0.5 m/s 2.0 kg 2.0 kg Assess: For an object with positive velocity, a negative impulse slows down an object and a positive impulse increases speed. The opposite is true for an object with negative velocity. P9.39.!Prepare:! Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We will also use constant-acceleration kinematic equations. Solve: To find the ball’s velocity just before and after it hits the floor: v12y v02y 2a y ( y1 y0 ) 0 m 2 /s 2 2( 9.8 m/s 2 )(0 2.0 m) v1 y 6.261 m/s v32y v22 y 2a y ( y3 y2 ) 0 m 2 /s 2 v22 y 2( 9.8 m/s 2 )(1.5 m 0 m) v2 y 5.422 m/s The force exerted by the floor on the ball can be found from the impulse-momentum theorem: mv2 y mv1 y area under the force curve (0.2 kg)(5.422 m/s) (0.2 kg)(6.261 m/s) 12 Fmax (5 103 s) Fmax 940 N Assess: A force of 940 N exerted by the floor is typical of such collisions. P9.47. Prepare:! Let the system be ball + racket. During the collision of the ball and the racket, momentum is conserved because all external interactions are insignificantly small. We will also use the momentum-impulse theorem. Solve: (a) The conservation of momentum equation ( px)f ( px)i is mR (vRx )f mB(vBx )f mR (vRx )i mB(vBx )i (1.0 kg)(vRx )f (0.06 kg)(40 m/s) (1.0 kg)(10 m/s) (0.06 kg)( 20 m/s) (vRx )f 6.4 m/s (b) The impulse on the ball is calculated from ( pBx )f ( pBx )i J x as follows: (0.06 kg)(40 m/s) (0.06 kg)( 20 m/s) J x J x 3.6 N s Favg t 3.6 Ns 360 N 10 ms Assess: Let us now compare this force with the ball’s weight wB mB g (0.06 kg)(9.8 m/s 2 ) 0.588 N. Thus, Favg 610 wB. This is a significant force and is reasonable because the impulse due to this force changes the direction as well as the speed of the ball from approximately 45 mph to 90 mph. Favg