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Capt’n Jack’s Momentum Calculations
1. Capt’n Jack pushes Elizabeth Swann into the soldiers. The resultant force on her is 10 N and
he pushes for 2 seconds. Her mass is 60 Kg. What is her momentum?
2x10 = 20 kg m/s
2. Capt’n Jack holds onto the rope.
a. The cannon falls for 5 seconds and experiences a change of momentum of 1,600 Kg
m/s. Calculate the resultant force.
b. When he reaches the top of the post, Jack has a velocity of 20 m/s. What is his
mass?
1600/5 = 320 N
1600/20 = 80 kg
3. Capt’n Jack then spins around the top of the post on a beam. The velocity of Jack and the
beam (just before he lets go) is 5 m/s. The beam has a mass of 150 Kg.
a. What is the momentum of Jack and the beam?
b. As Jack lets go of the beam, what is his momentum? (Assume that his speed doesn’t
change)
c. What is the momentum of the beam?
(150+80) x 5 = 1150 kg m/s
80 x 5 = 400 kg m/s
150 x 5 = 750 kg m/s
4. Jack slides down the rope. It takes him 6 seconds. The forward force acting on him during
this time is 20 N, the backwards force is 7 N.
a. Calculate his change in momentum.
b. Assuming that Jack was stationary when he started sliding, what is his velocity at the
bottom of the rope?
(20-7) x 6 = 78 kg m/s
78/80 = 0.98 m/s
5. The soldiers fire bullets at Capt’n Jack as he makes his escape. The mass of the bullet is 10 g,
the mass of the gun is 1 kg, and the momentum of the bullet is 6 kg m/s.
a. What is the momentum of the gun?
6 kg m/s
b. What is the velocity of the gun as it recoils?
6/1 = 6 m/s
c. Assuming that the recoil lasts for 0.5 seconds, what is the resultant force that the
gun exerts on the soldier?
6/0.5 = 12 N
6. In the blacksmiths workshop the change in momentum of the hammer is 10 kg m/s. It falls
for 1 second. What force does it exert on the chain?
Name______________________________________________________
10/1 = 10 N
7. Will Turner throws the sword at the door. His backwards velocity is 0.1 m/s, his mass is 70
kg. The mass of the sword is 1.2 kg.
a. What speed is the sword travelling at when it hits the door? (Give your answer to 1
d.p.)
b. When the sword hits the door it takes 0.2 s to stop moving. Calculate the resultant
force when the sword hits the door.
c. Jack pulls at the sword with a force of 20 N. Suggest why the sword doesn’t move.
70 x 0.1 = 7, 7/1.2 = 5.8 m/s
5.8/0.2 = 29 N
Jack has not overcome the forces of air resistance and friction.
8. When Jack falls on the barrow he takes 0.1 s to stop. The barrow transfers all of Jacks
momentum to Will. When Jack hits the barrel his velocity is 1 m/s. What is:
a. Will’s velocity upwards?
b. The force acting on Jack when he lands?
1 x 80 = 80, 80/70 = 1.14 m/s
80/0.1 = 800 N
9. When Will cuts down the sack it takes 2 s to fall, with a resultant force acting on it of 60 N.
a. What is its change in momentum?
b. When the sack hits the cart it takes 0.1 s to stop. Calculate the force that it transfers
to Jack.
c. This force acts on Jack for 0.1 s. What momentum does Jack have?
d. What velocity does Jack have?
e. Whose upwards velocity was fastest, Jack’s or Will’s (from question 8)?
60 x 2 = 120 kg m/s
120/0.1 = 1200 N
1200 x 0.1 = 120 kg m/s
120/80 = 1.5 m/s
Jack’s
10. The dog has a mass of 7 kg and a momentum of 14 kg m/s. The keys to the cells weigh 0.5
kg. When the dog picks up the keys (to go and taunt the pirates with), how much does his
velocity change? (Give your answer to 2 d.p.)
Before = 14/7 = 2.00 m/s
After = 14/7.5 = 1.87 m/s
Difference = 2.00 – 1.87 = 0.13 m/s