Solutions to HW14, Chapter 8 NOTE! The problems in masteringphysics.com had their numbers altered slightly for each individual student. The solutions below use the same numbers as those used in the book for that problem!

8.37. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic equations. Visualize:

Solve: As the rocket is accidentally bumped 0 0 5 m/sxv = . and 0 0 m/syv = . On the other hand, when the engine is fired

220 N 40 m/s0 500 kg

xx x x

FF ma a

m= ⇒ = = =

.

(a) Using 211 0 0 1 0 1 02( ) ( ) ,y yy y v t t a t t= + − + −

2 211 120 m 40 m 0 m ( 9 8 m/s ) 2 857 st t= + + − . ⇒ = .

The distance from the base of the wall is 21

1 0 0 1 0 1 02( ) ( )x xx x v t t a t t= + − + − 2 2120 m (0 5 m/s)(2 857 s) (40 m/s )(2 857 s) 165 m= + . . + . =

(b) The x- and y-equations are 2 21

0 0 0 022 21

0 0 0 02

( ) ( ) 40 4 9

( ) ( ) 0 5 20

y y

x x

y y v t t a t t t

x x v t t a t t t t

= + − + − = − .

= + − + − = . +

Except for a brief interval near 20, 20 0 5t t t= . .? Thus

220 ,x t≈ or 2 /20t x= . Substituting this into the y-equation gives

40 0 245y x= − .

This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.

8.40. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:

Note that we need to use the coefficient of static friction s,µ which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is

2

s cos sinrmvF f nr

θ θ∑ = + = s Gcos sin 0zF n f Fθ θ∑ = − − = N

Maximum speed is when the static friction force reaches its maximum value s max s( )f nµ= . Then 2

s( cos15 sin15 )mvnr

µ °+ ° =

s(cos15 sin15 )n mgµ°− ° =

Dividing these two equations and simplifying, we get 2

s s

s s

2

tan15 tan151 tan15 1 tan15

(1 0 0 268)(9 80 m/s )(70 m) 34 m/s(1 0 268)

v v grgr

µ µµ µ+ ° + °= ⇒ =

− ° − °

. + .= . =− .

Assess: The above value of 34 m/s 70 mph≈ is reasonable.

8.59. Model: Model the ball as a particle undergoing circular motion in a vertical circle. Visualize:

Solve: Initially, the ball is moving in circular motion. Once the string breaks, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile, which we can find by using the kinematic equation

2 2 2 2 2 21 0 1 0 0 02 ( ) 0 m /s ( ) 2( 9 8 m/s )(4 0 m 0 m) 8 85 m/syv v a y y v v= + − ⇒ = + − . . − ⇒ = .

This is the speed of the ball as the string broke. The tension in the string at that instant can be found by using the r-component of the net force on the ball:

rF∑ =20 yvT mr

⎛ ⎞⎜ ⎟= ⇒⎜ ⎟⎝ ⎠

2(8 85 m/s)(0 100 kg) 13 N0 60 m

T .= . =.

8.63. Model: Assume the particle model for a ball in vertical circular motion. Visualize:

Solve: (a) Newton’s second law in the r- and t-directions is

2

net( ) cos tr r

mvF T mg mar

θ= + = = net( ) sint tF mg maθ= − =

Substituting into the r-component, 2

2(20 N) (2 0 kg)(9 8 m/s )cos30 (2 0 kg)(0 80 m)

tv+ . . ° = ..

3 85 m/stv⇒ = .

The tangential velocity is 3.8 m/s. (b) Substituting into the t-component,

2(9 8 m/s )sin30 ta− . ° = 24 9 m/sta⇒ = − . The radial acceleration is

2 22(3 85 m/s) 18 5 m/s

0 80 mt

rvar

.= = = ..

Thus, the magnitude of the acceleration is 2 2 2 2 2 2 2 2(18 5 m/s ) ( 4 9 m/s ) 19 1 m/s 19 m/sr ta a a= + = . + − . = . ≈

The angle of the acceleration vector from the r-axis is

1 1 4 9tan tan 14 8 1518 5

t

r

aa

.= = = . ° ≈ °.

2 2o

The angle is below the r-axis.

8.66. Model: Treat Sam as a particle. Visualize: This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. Sam’s final velocity at the top of the slope is his initial velocity as he becomes airborne.

Solve: Sam’s acceleration up the slope is given by Newton’s second law:

net 0

2 20

( ) sin10200 Nsin10 (9 8 m/s )sin10 0 965 m/s75 kg

sF F mg maFa gm

= − ° =

= − ° = − . ° = .

The length of the slope is 1 (50 m)/sin10 288 ms = ° = . His velocity at the top of the slope is 2 2 21 0 0 1 0 0 1 12 ( ) 2 2(0 965 m/s )(288 m) 23 6 m/sv v a s s a s v= + − = ⇒ = . = .

This is Sam’s initial speed into the air, giving him velocity components 1 1cos10 23 2 m/sxv v= ° = . and

1 1sin10 410 m/syv v= ° = . This is not projectile motion because Sam experiences both the force of gravity and the thrust of his skis. Newton’s second law for Sam’s acceleration is

2net1

2net 2

1

( ) (200 N)cos102 63 m/s

75 kg

( ) (200 N)sin10 (75 kg)(9 80 m/s )9 34 m/s

75 kg

xx

yy

Fa

m

Fa

m

°= = = .

°− .= = = − .

The y-equation of motion allows us to find out how long it takes Sam to reach the ground:

2 2 212 1 1 2 1 2 2 220 m 50 m (4 10 m/s) (4 67 m/s )y yy y v t a t t t= = + + = + . − .

This quadratic equation has roots 2 2 86 st = − . (unphysical) and 2 3 74 st = . . The x-equation of motion—this time with an acceleration—is

2 2 21 12 1 1 2 1 2 2 22 20 m (23 2 m/s) (2 63 m/s ) 105 mx xx x v t a t t t= + + = + . − . =

Sam lands 105 m from the base of the cliff.