Physics 110 Homework Solutions Week #6 - Wednesday

Transcription

1 Physics 110 Homework Solutions Week #6 - Wednesday Friday, May3, 2013 Chapter 6 Questions - none Multiple-Choice 66 C 67 D 68 B 69 C Problems 612 It s velocity as the ball hits the ground is found from v fy = v 2 iy 2gΔy = 2 98 m 1m = 443 m s 2 s vertically down The momentum is vertically down and has a magnitude of p y = mv y = 05kg 443 m s = 221 kgm s 613 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 Δp x = p fx p ix = mv mv = 2mv = 2 01kg 5 m s away from the wall b The average force is F = Δp Δt kjgm 1 s = 0005s = 1 kgm s = 200N away from the wall or, 1 kgm/s directed 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 Δp wall + Δp ball = 0 and Δp wall = -Δp ball ; but the M wall is so large that the v wall is negligible 614 First we convert the speed from mph to m/s and 125 miles per hour translates to 556m/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 617 A railroad car a Using conservation of momentum, the final velocity is given by 10,000kg(24m/s) + 0 = (10,000kg+1,200kg)v final or v final = 214 m/s in the direction the railroad car was traveling b KE init = ½ (10,000)(24) 2 = 288 x 10 6 J KE final = ½ (11,200)(214) 2 = 256 x 10 6 J, so the % loss is [KE init KE final ]/KE init x 100 = 111%

2 c Frictional force = µ k F N = 09(11,200)(98) = 988 x 10 4 N d Work by friction = ΔKE = 0 ½ mv 2 = -1/2 (11,200)(214) 2 = x 10 6 J 618 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 the right directed to 624 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 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 to

3 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 615 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 2 nd quadrant Taking the ratio of these two equations produces an angle of 332 o above the x axis Then using any one of the above equations we find for the magnitude of the velocity to be 31m/s 619 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, v 2 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

4 velocity is energy lost is The fraction of the initial 620 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 energy lost is So the percent of the initial kinetic

5 Using the mass of an alpha particle of 4u and of gold 197u, we have 621 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 625 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 v ix in terms of v so that we can determine the unknown angle θ Conservation of energy gives Therefore we have

6 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 = 45 o