Mechanics Homework, Set 8

1. The linear momentum of a runner in a 100 m dash is $7.6*10^2$ KG x (m/s) If the runner’s speed is 13 m/s , what is his mass? Express answer using two significant figures.

2. A 16.0 g rubber bullet hits a wall with a speed of 180 m/s.If the bullet bounces straight back with a speed of120 m/s what is the magnitude of the change in momentum of the bullet? B) What is the direction of the change in momentum of the bullet?

3. A 0.22 kg billiard ball traveling at a speed of 16 m/s strikes the side rail of a pool table at an angle of 60 degrees. If the ball rebounds at the same speed and angle, what is the magnitude of the the change in its momentum? Express your answer using two significant figures. B) What is the direction of the the change in its momentum?

4. An automobile with a linear momentum of $3.1*10^4$ kg x (m/s) is brought to a stop in 5.5 s

What is the magnitude of the average braking force?

5. At a shooting competition, a contestant fires and a 13.0 g bullet leaves the rifle with a muzzle speed of 130 m/s. The bullet hits the thick target backing and stops after traveling 4.10 cm.

A) Assuming a uniform acceleration, what is the magnitude of the impulse on the target? B) What is the direction of the impulse on the target? C)What is the magnitude of the average force on the target? D)What is the direction of the average force on the target?

1.

by definition momentum = mass*speed

$p= m*v$

$m= p/v = 7.6*100/13 =760/13 =58.46 kg$

2.

Change of momentum $= p2-p1 = m(v2-v1) = m(120-(-180)) =16*10^{-3}*300 =4.8 kg*m/s$

taking the positive x axe back from the wall the direction of the change is of the momentum is in the positive x way (straight back from the wall).

3. taking the angle between the side rail and the incoming path being 60 degree only the normal (perpendicular) to the rail momentum changes by bouncing (the parallel to the rail momentum is the same).

change of momentum $= p2 _n-p1 _n = p2*sin(\alpha) -p1*sin(\alpha) =$

$= m v*sin(\alpha)-m*(-v)*sin(\alpha) = 2 m v*sin(\alpha) = 2*16*0.22*sin(60) =6.10 kg*m/s$

(if the angle is formed by the path and the normal to the rail one must simply put above the cos(60) instead the sin(60))

the direction is perpendicular to the rail, away from the rail.

4.the force is equal to the change ($\Delta$) in momentum over the time.

$F = \Delta(P)/\Delta(t) = 3.1*10^4/5.5 =5636.36 N$

5. One must assume that the bullet hits the target perpendicular.

a) All the momentum of the bullet at the output of the rifle is lost on to the target.

$P = m*v = 0.013*130 =1.69 kg*m/s$

b) the direction of the impulse on the target is perpendicular to the target, into the target.

c) $0 = v^2-2*a*d$, where a is the acceleration and v the speed of the bullet when hits the target

$a = v^2/2/d =130*130/2/0.041 =206097.56 m/s^2$

the force on to the target is equal and contrary to the force on the the bullet

$F = m*a =0.013 *206097.56 = 2679.27 N$

d) The direction of the average force on the target is perpendicular to the target into the target.