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What is the momentum of a 1000 kg motorbike travelling at 10 m/s?
p = m•v
p = 100 • 10
p = 1×103 kg•m/s
What is the mass of plane that is travelling at 200 km/h and has a momentum of 1.1 × 106 kg•m/s? (Step 1)
p = m•v
m = p/v
Velocity
200 km/h ÷ 3.6 = 55.6 m/s
What is the mass of plane that is travelling at 200 km/h and has a momentum of 1.1 × 106 kg•m/s? (Step 2)
m = p÷v
m = 1.1 × 106 kg•m/s ÷ 55.6 m/s
m = 2.0 × 104 kg
How fast does a 0.01 kg bug have to fly to have a momentum of 0.25 kg•m/s? Is this possible? (Step 1)
p = m•v
v = p/m
How fast does a 0.01 kg bug have to fly to have a momentum of 0.25 kg•m/s? Is this possible? (Step 2)
v = p÷m
v = 0.25/0.01
v = 25 m/s
Is it possible? No
A 3.95-kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of the rocket after the fuel has burned? (Ignore effects of gravity and air resistance)
m1v1 + m2v2 = m1v1’ + m2v2’
0 = m1v1’ + m2v2’
-m2v2’ = m1v1’
-0.05 • (-625) = 3.95 • v1’
31.25 = 3.95 • v1’
31.25/3.95 = v1’
+7.9 m/s = v1’
A thread holds two carts together on a frictionless surface. A compressed spring acts upon the carts. After 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?
0 = m1v1’ + m2v2’
-m2v2’ = m1v1’
4.5 • v2’ = 1.5 • (27)
v2’ = 40.5/4.5
v2’ = 9 cm/s
Two campers dock a canoe. One camper steps onto the dock. This camper has 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?
0 = m1v1’ +
-m1v1’ = m2v2’
-80 • 4.0 = 110 • v2’
-320 = 110 • v2’
v2’ = -2.9 m/s
Jenny has a mass of 35.6 kg and her skateboard has a mass of 1.3 kg. What is the momentum of Jenny and her skateboard together if they are going 9.50 m/s?
p=(m1+m2)v
p = (35.6+1.3) 9.50
p = 3.5 × 102 kg•m/s
A hockey player makes a slap shot, exerting a force of 30.0 N on the hockey puck for 0.16 s. What impulse is given to the puck?
FNet•△t = I
30 • 0.16 = I
4.8 N•s = I
The hockey puck shot in Problem 2 has a mass of 0.115 kg and was at rest before the shot. With what speed does it head toward the goal?
FNet•▵t = △p
FNet•▵t = m•▵v
FNet•▵t÷m = ▵v
4.8 ÷ 0.115 = ▵v
42 m/s = ▵v
A force of 6.00 N acts on a 3.00-kg object for 10.0 s.
a) What is the object’s change in momentum?
FNet•▵t = ▵p
6 • 10 = ▵p
+60 kg•m/s = ▵p
A force of 6.00 N acts on a 3.00-kg object for 10.0 s.
b) What is the change in velocity?
▵p = m•▵v
60 = 3.0•▵v
▵v = +20 m/s
The velocity of a 600-kg auto is changed from +10.0 m/s to +44.0 m/s in 68.0 s by an applied force, constant force.
a) What change in momentum does the force produce?
▵p = m•▵v
▵p = 600 • (44-10)
▵p = +2.0 × 104 kg•m/s
The velocity of a 600-kg auto is changed from +10.0 m/s to +44.0 m/s in 68.0 s by an applied force, constant force.
b) What is the magnitude of the force?
▵p = FNet•▵t
FNet = ▵p/▵t
FNet = 2.0 × 104 / 68
FNet = +3.0 × 102 N
A 845-kg drag race car accelerates from rest to 100 km/h in 0.90 s.
a) What is the change of momentum of the car?
Vf = 100 km/h ÷ 3.6 = 27.8 m/s
▵p = 845 • 27.8
▵p = +2.35 × 104 kg•m/s
A 845-kg drag race car accelerates from rest to 100 km/h in 0.90 s.
b) What average force is exerted on the car?
▵p = FNet•▵t
FNet = ▵p/▵t
FNet = 2.35 × 104 / 0.90
FNet = 2.6 × 104 N
A 95-kg fullback, 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?
p1 = m1 • v1
p1 = 95 • 8.2
p1 = +7.8 • 102 kg•m/s
A 95-kg fullback, 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.
b) What was the change in the fullback’s momentum?
▵p1 = p1f - p1i
▵p1 = 0 - 7.8 •102
▵p1 = -7.8 • 102 kg•m/s
A 95-kg fullback, 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.
c) What was the change in the fullback’s momentum?
▵p2 = -p1
▵p2 = - (-7.8 • 102)
▵p2 = 7.8 • 102 kg•m/s
A 95-kg fullback, 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.
d) What was the tackle’s original momentum?
▵p2 = p2f - p2i
+7.8 • 102 = 0 - p2i
p2i = -7.8 • 102 kg•m/s
A 95-kg fullback, 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.
e) How fast was the tackle’s original momentum?
p2 = m2v2
v2 = p2/m2
v2 = -7.8 • 102/128
v2 = -6.1 m/s
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)
I = -1.0 • 102 kg•m/s
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
b) -8 m/s
m • ▵v = m (Vf - Vi)
= 25 (-8-12)
= 25 (-20)
= -5.0 • 102 kg•m/s
A 2575-kg van runs into the back of a 825-kg compact car at rest. They move off together at 8.5 m/s. Assuming no friction with the ground, find the initial speed of the van.
m1v1 = (m1+m2) v’
2575 • v1 = (2575+825) • 8.5
2575 • v1 = (3400) • 8.5
2575 • v1 = 28900
v1 = 11.2 m/s
A 15-g bullet is shot into a 5085-g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a velocity of 1.0 m/s. Calculate the velocity of the bullet before the striking block.
m1v1 = (m1+m2) v’
15 • v1 = (15+5085) • 1.0
15 • v1 = 5100
v1 = 340 m/s
A hockey 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.
m1v1 = (m1+m2) v’
0.115 • 35.0 = (0.115 + 0.265) • v’
4.025 = (0.38) • v’
11 m/s = v’
A 50-kg woman, riding on a 10-kg cart, is moving east at 5.0 m/s. The woman jumps off the cart and hits the ground at 7.0 m/s eastward, relative to the ground. Calculate the velocity of the cart after she jumps off.
m1v+m2v = m1v’ + m2v2
v(m1+m2) = m1v1’ + m2v2’
5.0(50 + 10) = 50 • 7.0 + 10 • v2’
5.0 (60) = 350 + 10 • v2’
300 = 350 + 10 • v2’
-50 = 10 • v2’
-5 m/s = v2’
Two students on roller skates stand face-to-face, then push each other away. One student has a mass of 90 kg, the other 60 kg. Find the ratio of their velocities just after their hands lost contact. Which student has the greater speed?
(Note: the ratio would be -60/90 or -0.66… The 60 kg kid would have the greatest speed.
A car with mass 1245 kg, moving at 29 m/s, strikes a 2175-kg car at rest. If the two cars stick together, with what speed do they move?
m1v1 = (m1+m2) v’
1245 • 29 = (1245 + 2175) v’
36105 = (3420) v’
11 m/s = v’
