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1. A pitcher claims he can throw a 0.145 kg baseball with as much momentum as a speeding bullet. Assume that a 3.00 g bullet moves at a speed of 1.50 x 10^3 m/s.

a. What must the baseball's speed be if the pitcher's claim is valid?
B. Which has greater kinetic energy, the ball or the bullet?

2. What is the relationship between impulse and momentum?

3. A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocery cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and bag?

4. During practice, a student kicks a 0.40 kg soccer ball with a velocity of 8.5 m/s to the south into a 0.15 kg bucket lying on its side. The bucket travels with the ball after the collision.
a. What is the final velocity of the combined mass?

5. A 16.0 kg canoe moving to the left at 12 m/s makes an elastic head-on collision with a 4.0 kg raft moving to the right at 6.0 m/s. After the collision, the raft moves to the left at 22.7 m/s. Disregard any effects of the water.
A. Find the velocity of the canoe after collision.

https://brainmass.com/physics/conservation-of-momentum/momentum-kinetic-energy-impulse-11401

SOLUTION This solution is FREE courtesy of BrainMass!

1. A pitcher claims he can throw a 0.145 kg baseball with as much momentum as a speeding bullet. Assume that a 3.00 g bullet moves at a speed of 1.50 x 10^3 m/s.

a. What must the baseball's speed be if the pitcher's claim is valid?

momentum =mv

momentum of speeding bullet= 4.5 kg-m/s =3*10^-3*1.5*10^3

momentum of baseball=mv= 4.5 kg-m/s

therefore v= 31.03 m/s

baseball's speed= 31.03 m/s

B. Which has greater kinetic energy, the ball or the bullet?

KE= p^2/(2m) where p is the momentum= 4.5 kg-m/s for both bullet and the baseball

mass of bullet= 0.003 kg
Therefore KE= 3375 Joules

mass of bullet= 0.145 kg
Therefore KE= 69.83 Joules

Bullet has the greater KE as with the same momentum it has the smaller mass.

2. What is the relationship between impulse and momentum?

The time integral of the net force acting on a body is called the impulse which is equal to the change in momentum 3. A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocery cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and bag?

mass of bag of rice=m1= 9 kg
mass of grocery cart=m2= 18 Kg
Velocity v1= 5.5 m/s

Initial momentum=m1v1= 49.5 kg-m/s

Final momentum = (m1+m2) V as both the cart and the bag will move with the same velocity V

From the conservation of linear momentum since no external force is acting
Initial momentum=m1v1= 49.5 =Final momentum = (m1+m2) V

Therefore V= 1.83 m/s

4. During practice, a student kicks a 0.40 kg soccer ball with a velocity of 8.5 m/s to the south into a 0.15 kg bucket lying on its side. The bucket travels with the ball after the collision.
a. What is the final velocity of the combined mass?

mass of ball=m1= 0.4 kg
mass of bucket=m2= 0.15 Kg
Velocity v1= 8.5 m/s

Initial momentum=m1v1= 3.4 kg-m/s

Final momentum = (m1+m2) V as both the ball and the bucket will move with the same velocity V

From the conservation of linear momentum since no external force is acting
Initial momentum=m1v1= 3.4 =Final momentum = (m1+m2) V

Therefore V= 6.18 m/s

5. A 16.0 kg canoe moving to the left at 12 m/s makes an elastic head-on collision with a 4.0 kg raft moving to the right at 6.0 m/s. After the collision, the raft moves to the left at 22.7 m/s. Disregard any effects of the water.
A. Find the velocity of the canoe after collision.

mass of canoe=m1= 16 kg
mass of raft=m2= 4 Kg
Velocity v1(i) of canoe= -12 m/s right direction is positive
Velocity v2(i) of raft= 6 m/s right direction is positive

Initial momentum=m1v1(i)+ m2v2(i)= -168 kg-m/s

Velocity v1(f) of canoe= ? m/s
Velocity v2(f) of raft= -22.7 m/s right direction is positive

Final momentum=m1v1(f)+ m2v2(f)=

From the conservation of linear momentum since no external force is acting
Initial momentum= -168 kg-m/s =Final momentum =m1v1(f)+ m2v2(f)

m2v2(f)= -90.8 kg-m/s
Therefore m1v1(f)= -77.2 kg-m/s

or v1(f)= -4.825 m/s
Since the right direction is positive
Final velocity of canoe= 4.825 m/s to the left

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