Physics 4A Chapter 11 HW Solutions

Chapter 11 Conceptual Questions: 3, 6, 13 Exercises & Problems: 13, 21, 25, 41, 47, 54, 57, 71, 81

CQ 11.3. The impulse is equal to the change in momentum, so

f i 4 Nsx x xp mv mv∆ = − = −

The final velocity is thus

f i4 Ns 4 Ns1 m/s 1 m/s

2 kgx xv vm

= − = − = −

Since the velocity is negative, the object is now moving to the left with a speed of 1 m/s. Note that the impulse was negative, which decreases the initially positive velocity.

CQ 11.6. In this story, Carlos is correct. During the short time of the bullet-block collision other forces are negligible compared to the force between the bullet and block, so in the impulse approximation momentum is conserved. When the bullet bounces off of the steel block, the bullet’s final momentum is backward. To balance that, the steel block must be moving forward faster than the case in which the bullet embeds itself in the wooden block (in which case the bullet has a final momentum in the forward direction).

CQ 11.13. (a) Let Paula and Ricardo be a system. Initially, their total momentum is zero since they are at rest. After they push off each other, the total momentum must still be zero, so Ricardo and Paula must have equal but opposite momenta. (b) Momentum .p mv= Since Paula is less massive than Ricardo, her speed must be higher than Ricardo’s for her to have the same momentum as Ricardo.

11.13. Model: Model the ball as a particle, and its interaction with the wall as a collision in the impulse approximation. Visualize: Please refer to Figure EX11.13. Solve: Using the equations

f ix x xp p J= + and f

i( )

t

x xtJ F t dt= =∫ area under force curve

f

f

(0 250 kg) (0 250 kg)( 10 m/s) (500 N)(8 0 ms)

4.0 N( 10 m/s) 6.0 m/s0.250 kg

x

x

v

v

. = . − + .

= − + =

Assess: The ball’s final velocity is positive, indicating it has turned around.

11.21. Model: This is a two-part problem. First, we have an inelastic collision between Fred (F) and Brutus (B). Fred and Brutus are an isolated system. The momentum of the system during collision is conserved since no significant external force acts on the system. The second part involves the dynamics of the Fred + Brutus system sliding on the ground. Visualize:

Note that the collision is head-on and therefore one-dimensional. Solve: The equation f ix xp p= gives

F B f F i F B i B f

f

( ) ( ) ( ) (60 kg 120 kg) (60 kg)( 6.0 m/s) (120 kg)(4.0 m/s)0.667 m/s

x x x x

x

m m v m v m v vv

+ = + ⇒ + = − +

= The positive value indicates that the motion is in the direction of Brutus. The model of kinetic friction yields:

k k k F B F B k( ) ( ) x xf n m m g m m a a gµ µ µ= − = − + = + ⇒ = − Using the kinematic equation 2 2

1 0 1 02 ( ),x x xv v a x x= + − we get 2 2 2 2 2 21 0 k 1 f 1

2 2 2 21 1

2 0 m /s 2(0 30)(9 8 m/s )

0 m /s (0 667 m/s) (5 9 m/s ) 7.6 cmx x xv v g x v x

x x

µ= − ⇒ = − . .

= . − . ⇒ = They slide 7.6 cm in the direction Brutus was running. Assess: After the collision, Fred and Brutus slide with a small speed but with a good amount of kinetic friction. A stopping distance of 7.6 cm is reasonable.

11.25. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision. Visualize:

Solve: For a package with mass m the conservation of energy equation is

2 21 g1 0 g0 1 1 0 0

1 1( ) ( )2 2m mK U K U m v mgy m v mgy+ = + ⇒ + = +

Using 0 1( ) 0 m/s and 0 m,mv y= =

2 21 0 1 0

1 ( ) ( ) 2 2(9 8 m/s )(3 0 m) 7 668 m/s2 m mm v mgy v gy= ⇒ = = . . = .

(a) For the perfectly inelastic collision the conservation of momentum equation is

f i 2 3 1 1 2( 2 )( ) ( ) (2 )( )x x m m mp p m m v m v m v= ⇒ + = +

Using 1 2( ) 0 m /s, we getmv =

2 3 1( ) ( ) /3 2 56 m/sm mv v= = .

The packages move off together at a speed of 2.6 m/s. (b) For the elastic collision, the mass m package rebounds with velocity

3 12 1( ) ( ) (7.668 m/s) 2.56 m/s2 3m m

m mv vm m

−= = − = −

+ The negative sign with 3( )mv shows that the package with mass m rebounds and goes to the position 4.y We can determine 4y by applying the conservation of energy equation as follows. For a package of mass m:

2 2f gf i gi 4 4 3 3

1 1( ) ( )2 2m mK U K U m v mgy m v mgy+ = + ⇒ + = +

Using 3 3 4( ) 2 55 m/s, 0 m, and ( ) 0 m/s,m mv y v= − . = = we get

24 4

1 ( 2 56 m/s) 33 cm2

mgy m y= − . ⇒ =

11.41. Model: Model the ball as a particle that is subjected to an impulse when it is in contact with the floor. We shall also use constant-acceleration kinematic equations. During the collision, ignore any forces other than the interaction between the floor and the ball in the impulse approximation. Visualize:

Solve: To find the ball’s velocity just before and after it hits the floor: 2 2 2 2 21 0 1 0 12 2 2 2 2 23 2 3 2 2 2

2 ( ) 0 m /s 2( 9.8 m/s )(0 2.0 m) 6.261 m/s

2 ( ) 0 m /s 2( 9.8 m/s )(1.5 m 0 m) 5.422 m/s

y y y y

y y y y y

v v a y y v

v v a y y v v

= + − = + − − ⇒ = −

= + − ⇒ = + − − ⇒ =

The force exerted by the floor on the ball can be found from the impulse-momentum theorem:

2 1 1 area under the force curvey y ymv mv Fdt mv= + = +∫

31max2

2max

(0.200 kg)(5.422 m/s) (0.200 kg)(6.261 m/s) (5.0 10 s)

9.3 10 N

F

F

−= − + ×

= ×

Assess: A maximum force of 29.3 10 N× exerted by the floor is reasonable. This force is the same order of magnitude as the force of the racket on the tennis ball in the Problem 11.38.

11.47. Model: This problem deals with a case that is the opposite of a collision. Our system comprises of three coconut pieces that are modeled as particles. During the explosion, the total momentum of the system is conserved in the x- and y-directions.

Visualize:

Solve: The initial momentum is zero. From f i ,x xp p= we get

1 f 1 0 01 f 1 3 f 3 f 3

3

( ) ( )( ) ( ) cos 0 kg m/s ( ) cos2 2

m v m v vm v m v vm m

θ θ − − −+ + = ⇒ = = =

From f i ,x xp p= we get

2 f 2 0 02 f 2 3 f 3 f 3

3

( ) ( )( ) ( ) sin 0 kg m/s ( ) sin2 2

m v m v vm v m v vm m

θ θ − − −+ + = ⇒ = = =

2 210 0 0

f 3( ) , tan (1) 452 2 2v v vv θ − = + = = = °

The speed to the third piece is 14 m/s at 45° east of north.

11.54. Model: Model the two blocks (A and B) and the bullet (L) as particles. This is a two-part problem. First, we have a collision between the bullet and the first block (A). Momentum is conserved since no external force acts on the system (bullet + block A). The second part of the problem involves a perfectly inelastic collision between the bullet and block B. Momentum is again conserved for this system (bullet + block B). Visualize:

Solve: For the first collision the equation f ix xp p= is

L 1 L A 1 A L 0 L A 0 A( ) ( ) ( ) ( )x x x xm v m v m v m v+ = + 1 L 1 L(0.010 kg)( ) (0.500 kg)(6.0 m/s) (0.010 kg)(400 m/s) 0 kg m/s ( ) 100 m/sx xv v+ = + ⇒ =

The bullet emerges from the first block at 100 m/s. For the second collision the equation f ix xp p= is

L B 2 L 1 L 2 2( ) ( ) (0 010 kg 0 500 kg) (0 010 kg)(100 m/s) 2.0 m/sx x x xm m v m v v v+ = ⇒ . + . = . ⇒ =

11.57. Model: The track is frictionless. Visualize: cv Rg= Solve: (a) First use conservation of momentum during the collision, then conservation of energy as the combined block goes to the top of the loop.

i fp p∑ = ∑

tot0 ( )mmv m M v+ = +

totmm Mv v

m+

=

Now use the conservation of energy. totv is the speed of the combined block just after the collision.

i i f fU K U K+ = + 2 2tot c

1 10 ( ) ( ) (2 ) ( )( )2 2

m M v m M g R m M v+ + = + + +

Cancel ( )m M+ and replace cv with .Rg

2tot tot4 5 5v Rg Rg Rg v Rg= + = ⇒ =

tot 5mm M m Mv v Rg

m m+ +

= =

(b) First use conservation of momentum and kinetic energy during the collision, then conservation of mechanical energy as the big block goes to the top of the loop. Call the speed of M just after the elastic collision V and the speed of m just after the collision .mv′

i fp p∑ = ∑

We drop the vectors because this is one-dimensional motion, but v′ may be negative.

0mmv mv MV′+ = +

mMv v Vm

′= +

Now use the conservation of energy as the block goes to the top of the loop.

i i f fU K U K+ = +

( )22 2 21 1 1 10 (2 )2 2 2 2m mmv MV Mg R mv M Rg′ ′+ + = + +

Subtract 21

2 mmv′ from both sides, cancel M, and solve for V.

4 5V Rg Rg Rg= + =

Now go back to the conservation of kinetic energy in the elastic collision.

i fK K∑ = ∑ 2 2 21 1 1

2 2 2m mmv mv MV′= +

Cancel 12 and divide by m.

2 2 5m mMv v Rgm

′= +

Find 2mv′ from the momentum equation.

22 5m m

M Mv v V Rgm m

= − +

22 2 2 5m m m

M M Mv v v V V Rgm m m

= − + +

Subtract 2mv from both sides, cancel ,M

m and solve for .mv

1 1 5 1 1 15 5 52 2 2 2 25m

M Rg M m Mv V Rg Rg Rgm m mRg

+= + = + =

Assess: We expected the initial speed needed to be greater for the inelastic case because kinetic energy isn’t conserved in the collision.

11.71. Model: Model the three balls of clay as particle 1 (moving north), particle 2 (moving west), and particle 3 (moving southeast). The three stick together during their collision, which is perfectly inelastic. The momentum of the system is conserved. Visualize:

Solve: The three initial momenta are

i1 1 i1ˆ ˆ(0 020 kg)(2.0 m/s) 0.040 kg m/sp m v j j= = . =

i2 2 i2ˆ ˆ(0 030 kg)( 3 0 m/s ) 0.090 kg m/sp m v i i= = . − . = −

i3 3 i3ˆ ˆ ˆ ˆ(0 040 kg)[(4 0 m/s)cos45 (4 0 m/s)sin45 ] (0 113 0 113 ) kg m/sp m v i j i j= = . . ° − . ° = . − .

Since f i i1 i2 i3,p p p p p= = + +

we have

1 2 3 f fˆ ˆ ˆ ˆ( ) (0 023 0 073 ) kg m/s (0 256 0 811 ) m/sm m m v i j v i j+ + = . − . ⇒ = . − .

2 2

f (0 256 m/s) ( 0.811 m/s) 0 85 m/sv = . + − = .

f1 1

f

0 811tan tan 720

below the -axi .256

sy

xx

v

vθ − − .

= = = °.

11.81. Model: The projectile + wood ball are our system. In the collision, momentum is conserved. Visualize:

Solve: The momentum conservation equation f ix xp p= is

P B f P i P B i B f i P( ) ( ) ( ) (1.0 kg 20 kg) (1.0 kg)( ) 0 kg m/sx x x x xm m v m v m v v v+ = + ⇒ + = + i P f( ) 21x xv v=

We therefore need to determine fx.v Newton’s second law for circular motion is 2

P B fG P B

( )( ) xm m vT F T m m gr

+− = − + =

Using max 400 N,T = this equation gives

2f

f max(1.0 kg 20 kg)400 N (1.0 kg 20 kg)(9.8 m/s) ( ) 4.3 m/s

2.0 mx

xv v+

− + = ⇒ =

Going back to the momentum conservation equation,

i f( ) 21 (21)(4 3 m/s) 90 m/sx P xv v= = . = That is, the largest speed this projectile can have without causing the cable to break is 90 m/s.