Momentum + Impulse Physics 11

Page 194 Question 16.

A 95-kg full back, running at 8.2 m/s, collides in midair with a 128-kg defensive tackle moving in the opposite direction. Both players end up with zero speed.

a) what was the fullback’s momentum before the collision?

m1 = 95kg v1 = 8.2 m/s v1’ = 0

m2 = 128kg v2 = ? v2’ = 0

m1v1 + m2v2 = m1v1’ + m2v2’

p = mv p = (95)(8.2) p= 779 or 7.8 × 10² kg*m/s

b) what is the change in the fullback’s momentum?

△p = F△t

△p = m (vt -vi) △p = 95 (0 - 8.2) △p = -799 N*S or -7.8 × 10² kg*m/s

c) What was the change in the tackle’s momentum?

△p ₂ = -△p₂ △p₂ = -(-7.8 × 10²)

△p₂ = 7.8 × 10² kg * m/s

d) what was the tackle’s original momentum?

△p ₂ = p₂‘ * p₂

7.8 × 10² = 0 - (-7.8 × 10²) = △p ₂ = 7.8 × 10² kg * m/s

18. before a collision, a 25-kg object is moving at +12 m/s. find the impulse that acted on this object if after the collision it moves at…

a) +8 m/s

m * △v = m (vf - vi)

= 25 (8 - 12)

= 25 ( -4 )

= -1.0 × 10² kg*m/s

b) -8 m/s

m * △v = m (vf - vi)

= 25 (-8 - 12)

= 25 ( -20 )

= -5.0 × 10² kg*m/s ← is negative because it is reversed in direction

19. A 2575-kg van runs into the back of a 825-kg compact car at rest. they move off together at 8.5m/s. assuming no friction with the ground, find the initial speed of the van.

m1 = 2575kg v1 = ? v1’ = 8.5 m/s

m2 = 825kg v2 = 0 v2’ = 8.5 m/s

m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 ( 0 ) = m1v1’ + m2v2’

2575vi = (2575) (8.5) + (825) (8.5)

= +11.2 m/s

= +11m/s

20. a 15-g bullet is shot into a 508-g wooden block, with the bullet in it, and acquires a velocity of 1.0m/s. calculate the velocity of the bullet before striking the block.

m1 = 15kg v1 = ? v1’ = 1.0 m/s

m2 = 5085kg v2 = 0 v2’ = 1.0 m/s

m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 ( 0 ) = m1v1’ + m2v2’

15v1 = (15) (1.0) + (5085) (1.0)

15v1 = 5100

divide both sides by 15

v1 = 340 m/s

v1 = 3.4 × 10² m/s

21. A hock puck, mass 0.115 kg, moving at 35.0 m/s, strikes a rubber octopus thrown on the ice by a fan. the octopus has a mass of 0.265 kg. the puck and octopus slide off together. find their velocity.

m1 = 0.115kg v1 = 35.0m/s v1’ = ?

m2 = 0.265kg v2 = 0 v2’ = ?

m1v1 + m2v2 = m1v1’ + m2v2’

(0.115) (35.0) + (0.265) (0) = (0.115)v1’ + (0.265)v2’

4.025 = (0.115)vi’ + (0.265)v2’

divide both sides by (0.115) + (0.265)

v’ = 10.6m/s

Pg 178-179

1. a compact car, mass 725 kg, is moving at +100km/h

a) find its momentum

b) at war velocity is the momentum of a ,arger car, mass 2175kg, equal to that of the smaller car?

a snowmobile has a mass of 2.50 × 10² kg. a constant force is exerted on it for 60.0s. the snowmoblie’s initial velocity is 6.00 m/s and its final velocity 28.0 m/s

a) what is its change in momentum?

b) what is the magnitude of the force exerted on it?

3. the brakes exert a 6.40 × 10² N force on a car weighting 15 680 N and moving at 20.0m/s. the car finally stops.

a) what is the cars mass

b) what is its initial momentum

c) what is the change in the cars momentum?

d) how long does the breaking force act on the car to bring it to a halt?

P 188

9. A 4.00-kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has burned? (Ignore effects of gravity and air resistance.)

10. A thread holds two carts together on a frictionless surface as in Figure 9-12. A compressed spring acts upon the carts. Atter the thread is burned, the 1.5-kg cart moves with a velocity of 27 cm/s to the left. What is the velocity of the 4.5-kg cart?

11. Two campers dock a canoe. One camper steps onto the dock. This camper has a mass of 80.0 kg and moves forward at 4.0 m/s. With what speed and direction do the canoe and the other camper move if their combined mass is 110 kg?