Physics 110 Homework Solutions Week #6 - Wednesday Friday, May3, 2013 Chapter 6 Questions - none

Multiple-Choice 6.6 C 6.7 D 6.8 B 6.9 C Problems 6.12 It’s velocity as the ball hits the ground is found from

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v fy = viy2 − 2gΔy = 2 × 9.8 m

s2×1m = 4.43 m

s vertically down. The momentum is

vertically down and has a magnitude of

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py = mvy = 0.5kg × 4.43 ms = 2.21 kgms .

6.13 A ball bouncing off of a wall

a. Assuming the positive x-direction is in the direction of the ball’s initial velocity into the wall, the change in momentum is

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Δpx = pfx − pix = −mv −mv = −2mv = −2 × 0.1kg × 5 ms = −1 kgms or, 1 kgm/s directed

away from the wall.

b. The average force is

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F =ΔpΔt

=−1 kjgms0.005s

= 200N away from the wall.

c. Yes it did, because the momentum of the (ball + wall) is conserved. The force on the wall from the ball is equal and opposite of the force on the ball from the wall and so Δpwall + Δpball = 0 and Δpwall = -Δpball; but the Mwall is so large that the vwall is negligible.

6.14 First we convert the speed from mph to m/s and 125 miles per hour translates to

55.6m/s. For a tennis ball launched from rest, the change in the momentum of the object is given as . By the impulse momentum theorem, (Newton’s second law) the average force is

.

6.17 A railroad car

a. Using conservation of momentum, the final velocity is given by 10,000kg(24m/s) + 0 = (10,000kg+1,200kg)vfinal or vfinal = 21.4 m/s in the direction the railroad car was traveling.

b. KEinit = ½ (10,000)(24)2 = 2.88 x 106 J KEfinal = ½ (11,200)(21.4)2 = 2.56 x 106 J, so the % loss is [KEinit – KEfinal]/KEinit x 100 = 11.1%

c. Frictional force = µkFN = 0.9(11,200)(9.8) = 9.88 x 104 N. d. Work by friction = ΔKE = 0 – ½ mv2 = -1/2 (11,200)(21.4)2 = - 2.56 x 106 J.

6.18 A roller coaster

a. Using conservation of energy between the initial point and point A we have the speed of the object as

,

where we take the zero of the gravitational potential energy to be at ground level. The centripetal force, directed vertically upward at point A, has magnitude

.

b. To determine the speed at point B we use conservation of energy between points A and B. We have

. c. Point C is at the zero of gravitational potential energy and given that energy is

conserved, the speed of the car at point C has to be the same as at point A or 14 m/s. Using conservation of momentum we have

directed to

the right. 6.24 A bullet and a block

a. Assuming that the positive x-direction is to the right we apply conservation of momentum. We find for the velocity after the collision

to

the right. b. Define d as the distance the block slides along the ramp and h as the height the

block rises above the horizontal, we have from the geometry

. Applying conservation of energy between the bottom of

the ramp and where the block comes to rest we have

c. In the presence of friction, energy is lost to heat between the surfaces. To calculate the new distance we use

Monday, May 6, 2013 Chapter 6 Questions - none Multiple-Choice - none Problems 6.15 At the highest point of the rocket’s motion, its velocity is zero. Therefore the

initial x- and initial y-momenta are both zero when the rocket explodes. After the explosion we apply conservation of momentum in the vertical and horizontal directions. Assuming that the piece of mass m has a momentum in the same quadrant as the 3m piece, we have in the vertical direction

. In the horizontal direction we have . Here we have that the x- component of the velocity is negative, while the y-component is positive, so the momentum vector lies in the 2nd quadrant. Taking the ratio of these two equations produces an angle of 33.2o above the –x axis. Then using any one of the above equations we find for the magnitude of the velocity to be 31m/s.

6.19 Using a standard Cartesian coordinate system we apply conservation of momentum in the horizontal and vertical directions and we have

and. Here we have two

equations and two unknowns, v2 and φ. Inserting the numbers from the problem we have for the horizontal and vertical directions

and . Dividing these two expressions we solve for the unknown angle and find

. Therefore the unknown

velocity is . The fraction of the initial energy lost is

6.20 The percent kinetic energy lost is given by

.

We need to determine both the final speed of the alpha particle and the gold nucleus after the collision. To do this we apply conservation of momentum and kinetic energy. From conservation of momentum we have

and from conservation of kinetic energy

. Here we have two equations and two unknowns. From momentum we solve for the final velocity of the alpha particle and obtain

. We square this result and substitute into the equation for

kinetic energy we obtain a quadratic equation in the final velocity of the gold

nucleus. The quadratic equation is . Using

the quadratic formula we find the solutions and reject the

zero speed solution. Substituting this result into our equation for the final speed of the alpha particle and we calculate the final speed to be

. So the percent of the initial kinetic

energy lost is

Using the mass of an alpha particle of 4u and of gold 197u, we have

.

6.21 The ballistic pendulum a. Conservation of momentum gives

.

b. Applying conservation of energy immediately after the collision we have and solving this

expression for the velocity of the ball and pendulum after the collision we have.

c. Substituting the expression in part b in to part a, we can calculate the initial velocity of the ball before the collision. We find

d. The fraction of the initial kinetic energy lost is

6.25 We break up the momentum into x and y-components and use conservation of

momentum in each direction. We have in the x-direction , while in the y-direction

. Using the results from the vertical motion we rewrite the x-momentum as

. Next we use the kinetic energy to obtain an expression for vix in terms of v so that we can determine the unknown angle θ. Conservation of energy gives

. Therefore we have

the magnitude of the velocity after the collision as . And from the x-

momentum we calculate the angle to be

. The angle of the velocity vector

after the collision is θ = φ = 45o.