Download Physics 192 Solutions to Mastering Physics Week 1

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  103 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 