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**Impulse** is the change in momentum of an object, and is a vector quantity: **j** = Δ**p**

**Momentum** is the product of the mass and the velocity of an object, and is a vector quantity:

**p** = m**v**

Since in daily life we do not notice an object's mass change just because its velocity changes, change in momentum is given by:

Δ**p** = mΔ**v**

**Impulse** occurs as force is applied to an object over a period of time: **j** = **F**Δt

Therefore, since **j** = Δ**p**

**F**Δt** = **mΔ**v**

**F**** = **mΔ**v**/Δt

**Newton's Second Law of Motion**: force is the rate of change of momentum,

F = Δ**p**/t

What is the momentum of an object with m=2.00 kg and v=40.0 m/s?

p = mv

p = (2.00)(40.0) = 80.0 kg*m/s

A force of 30000 N is exerted for 4.00 s, on a 95,000 kg mass.

(a) What is the impulse of the force for this 4.00 s?

(b) What is the mass's change in momentum from this impulse?

(c) What is the mass's change in velocity from this impulse?

(d) Why can't we find the resulting change in kinetic energy of the mass?

a. Impulse = Ft = 30000(4) = 120000 N-s

b. change in momentum = impulse = 120000 N-s

c. j = m∆v

120000 = (95000)∆v

∆v = 1.26 m/s

d. We do not know the initial velocity.

(See bottom of page for answers.)

1. A particle with momentum m**v**_{1} experiences a force which leaves it with momentum m**v**_{2}. Draw an arrow to represent the impulse the particle experienced.

2. A 1000 kg car accidentally drops from a crane and crashes at 30 m/s to the ground below and comes to an abrupt halt. What impulse acts on the car when it crashes?

3. If a force of 300N is exerted upon a 60 kg mass for 4 seconds, how much impulse does the mass experience?

4. Two goblins with identical mass are traveling on a frictionless surface at right angles to each other with velocities **v**_{1} and **v**_{2} respectively (**v**_{1} = 2**v**_{2}). The goblins collide and stick together.

a. Draw momentum vectors for both goblins before the collision.

b. Draw the momentum vector for the combined body immediately after the collision.

c. What is the impulse of the system?

5. A billiard ball approaches a cushioned edge of a billiard table with momentum, **p**. After the collision with the cushion, it bounces straight back with the same amount of momentum in the opposite direction. What is the impulse on the ball?

6. A trajectile, of mass 20 g, traveling at 350 m/s, strikes a steel plate at an angle of 30-degrees with a plane of the plate. It ricochets off at the same angle, at a speed of 320 m/s. What is the magnitude of the impulse that the steel plate gives to the trajectile?

7. A daredevil (m = 77.00 kg) tied to a 32.00 m bungee cord, leaps off a 62.00 m tall platform. He falls to 9.00 m above the ground before the bungee cord pulls him back up. What size impulse is exerted on the daredevil while the cord stretches.

8. An 80-kg man and his car are suddenly accelerated from rest to a speed of 5 m/s as a result of a rear-end collision. Assuming the time taken to be 0.3s, find the

a) impulse on the man and

b) the average force exerted on him by the back seat of his car

9. The momentum of a 30.0-g sparrow with a speed of 12 m/s is 0.36 kg*m/s. What will be its momentum 12s later if a constant .02 N force due to air resistance acts on it?

10. Two rickshaws, one twice the mass of the other, experience the same force for the same time. What is their difference in momentum? What is their difference in kinetic energy?

11. A racket exerted an average force of 152.0 N on a ball initially at rest. If the ball has a mass of 0.070 kg and was in contact with the racket for 0.030 s, what was the kinetic energy of the ball as it left the racket?

12. A 0.15-kg rubber ball's velocity just before impact with a floor is 6.5 m/s down, and just after is 3.5 m/s straight up. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?

Selected solutions are printed below.

For solutions to all the problems on this page click here.

1. **j** = m**v**_{2} - m**v**_{1} = m**v**_{2} + (-m**v**_{1})

2.

j = mΔv assuming constant mass.

j = 1000(-30) = -30 000 kg-m /s

**The impulse is 30000 kg-m/s upwards. **

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