Practice Problems (p.21-27)

To go from this…………………………...to this.

An object experiences a force of 19.97 N for a time period of 4.58 s. What is the impulse of the object?

Impulse

J=F( t)

Solution

J= 19.97 (4.58)

For what time must you exert a force of 45 N to get an impulse of 16 Ns?

Try another one!

16=45(x) X= .36

Solution

A 142.5 kilogram motorcycle is moving at a speed of 67.5 m/s. What is the momentum of the

cycle?

Momentum

SolutionP=mvP=142.5(67.5)

What is the momentum of a 23 Kg cannon shell going 530 m/s?

Try another one!

P=mvP=23(530)P= 12,190

Solution

Judy (mass=40.0 kg) is standing on slippery ice and catches her leaping dog, Atti (mass=15 kg),

moving horizontally at 3.0 m/s. What is the speed of Judy and her dog after the catch?

Conservation of Momentum

mJudyi(vJudyi)+ mAtti(vAtti)= mJudyf(vJudyf)+ mAttif(vAttif)

Solution

(40)(0) + (15)(3)==40

kg 15 kg

V= 3m/sP=45 (15x3)

V= 0 m/sP= 0

P=0+45P=45

55 kg

V= 45=55(v) .82 m/s P=45

A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After

the collision, both players move east at 2 m/s. What is their final momentum?

Try another one!

Solution80 kg 120 kg

V= 8 m/s P= 640

V= -2 m/s P=-240

P= 640+-240 =400

V=2 m/sP= 400

200 kg

Amplitude/ Period

Amplitude

Period

1. A force of 600 Newtons will compress a spring 0.5 meters. What is the spring constant

of the spring?

2. A spring has spring constant 0.1 m/Newton. What force is necessary to stretch the spring by

2 meters?

Hooke’s Law

1. 600= -.5 (x)=-1200

2. F= -.1(2)=-0.2

Solution FHooke’s=-k( x)

What is the force required to stretch a spring whose constant value is 100 N/m by an amount of 0.50 m?

Try another one!

FHooke’s=-k( x)

Solution

F= 100(.50)F=50 N

In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk.

The force you exerted on the door was 50N, applied perpendicular to the plane of the door. The door is 1.0m wide. Assuming that you pushed the door at

its edge, what was the torque on the swinging door (taking the hinge as the pivot point)?

Torque

SolutionT=(1.0m) (50N)T= 50Nm

A force of 20 N is applied perpendicular to the end of a bar of length 0.5 m. Calculate the torque

produced by the force.

Try another one!

T= (0.5)(20N)T=10Nm

Solution