NOTE: that i just want to use these practice problems to help for the
real test.  I just need step by step solving to get this right, and i
appreciate what you guys are doing thanks.

1 (part 1 of 2) 
A wooden block hangs on a string. A bullet
hits and sticks in the block.
What is conserved in the collision of the
bullet with the block?
1. Only mechanical energy
2. Momentum and mechanical energy
3. Only momentum
4. Neither momentum nor mechanical energy

2 (part 2 of 2) 
After the bullet collides and sticks in the
wooden block that hangs on a string, the
block swings out from its lowest position and
reaches its highest position.
What is conserved in the motion after the
bullet has collided with the block and the
block swings up to the highest position?
1. Momentum and mechanical energy
2. Neither momentum nor mechanical energy
3. Only mechanical energy
4. Only momentum

3 (part 1 of 4)
Assume: The positive direction is that of the
bullet.
A 28 kg gun is standing on a frictionless
surface. The gun fires a 54 g bullet with a
muzzle velocity of 301 m/s.
Calculate the momentum of the bullet immediately after the gun was fired. 

4 (part 2 of 4) 
Calculate the momentum of the gun immediately after the gun was fired.

5 (part 3 of 4) 10 points
Calculate the kinetic energy of the bullet immediately after the gun was fired. 

6 (part 4 of 4) 
Calculate the kinetic energy of the gun immediately after the gun was fired. 

7 (part 1 of 1) 
Two students on roller skates stand face-to-
face, then push each other away. One student
has a mass of 91 kg and the second student
55 kg.
Find the ratio of the magnitude of the first
student's velocity to the magnitude of the
second student's velocity.

8 (part 1 of 1) 
A hockey player makes a slap shot, exerting a
force of 38:4 N on the hockey puck for 0:114 s.
What impulse is given to the puck? 

9 (part 1 of 1)
A(n) 1160 kg car moving with a speed of
18:5 m/s collides with a utility pole and is
brought to rest in 0:13 s.
Find the magnitude of the average force exerted on the car during the collision. 

10 (part 1 of 1) 
A force of 1630 N is needed to bring a car
moving at 24:9 m/s to a halt in 15:3 s.
What is the mass of the car? 

11 (part 1 of 1) 
A(n) 16000 kg freight car is rolling along a
track at 3:8 m/s. Calculate the time needed for a force of
1050 N to stop the car.

Momentum is always conserved unless an outside force acts.
Mechanical energy is always conserved unless some of it is converted
to another form of energy, usually heat.


1 (part 1 of 2) 
A wooden block hangs on a string. A bullet
hits and sticks in the block.
What is conserved in the collision of the
bullet with the block?
1. Only mechanical energy
2. Momentum and mechanical energy
3. Only momentum
4. Neither momentum nor mechanical energy


Only momentum is conserved.  Much of the initial energy of the bullet
is converted to heat when it becomes lodged in the block.

2 (part 2 of 2) 
After the bullet collides and sticks in the
wooden block that hangs on a string, the
block swings out from its lowest position and
reaches its highest position.
What is conserved in the motion after the
bullet has collided with the block and the
block swings up to the highest position?
1. Momentum and mechanical energy
2. Neither momentum nor mechanical energy
3. Only mechanical energy
4. Only momentum


Only energy is conserved.  Actually, momentum is conserved too if you
include the entire earth in your system, because the force the string
applies to the earth actually does move it the tiniest of bits.  But
since our system is just the block, bullet, and string, momentum is
not conserved, because an outside force acts on the string during the
time the block is swinging.

3 (part 1 of 4)
Assume: The positive direction is that of the
bullet.
A 28 kg gun is standing on a frictionless
surface. The gun fires a 54 g bullet with a
muzzle velocity of 301 m/s.
Calculate the momentum of the bullet immediately after the gun was fired. 

Mb - mass of bullet (54 g)
Mg - mass of gun (28 kg)
M - mass of gun + bullet (28.054 kg)
Vb - velocity of bullet after firing (unknown)
Vg - velocity of gun after firing (unknown)
V - muzzle velocity of bullet (301 m/s)


Conservation of momentum:        Mb * Vb + Mg * Vg = 0
Definition of muzzle velocity:   Vb - Vg = V

Solving this system of two equations gives:

Vb = Mg * V / M
Vg = -Mb * V / M

The momentum of the bullet is Mb * Vb = MbMgV/M = 16.223 kg m/s
The momentum of the gun is Mg * Vg = -MbMgV/M = -16.223 kg m/s
Kinetic energy of bullet = .5 * Mb * Vb^2 = .5Mb(MgV/M)^2 = 2436.8 J
Kinetic energy of gun = .5 * Mg * Vg^2 = .5Mg(MbV/M)^2 = 4.6996 J

7 (part 1 of 1) 
Two students on roller skates stand face-to-
face, then push each other away. One student
has a mass of 91 kg and the second student
55 kg.
Find the ratio of the magnitude of the first
student's velocity to the magnitude of the
second student's velocity.

Momentum is conserved, so M1V1 + M2V2 = 0
The first student's speed is 55/91 as large as that of the second student.

For 8, 9, 10, and 11, the answers can be easily found from the relations:

Impulse = final momentum - initial momentum = force * time

you have to actually state the boundaries or limits set by the
questions in order to get the full answer. Most physics questions like
the ones you asked often come with statements that explicitly state
the questions limits. Such statements include:-
*neglect all outside forces
*assume no heat is generated and no energy is lost
and many more. Withouth these statements we can only assume that we
have to take into account EVERY acing force in the question and this
is both tedious and unnecessary.

When a bullet hits the block, what needs to be considered is wether
the collision is elastic in nature. If it is elastic, then the kinetic
energy is constant throughout the collision and vice-versa.
Realistically your collision will never be elastic since a lot of
energy is lost in the form of both heat and sound, however for
examination purposes sometimes they may ask you assume that it is
elastic. Momentum is always conserved unless an outside force is
present , in this case we can safely assume there is no such force.
However once the block begins to move , it will move in the circular
motion upward. Once it moves upward, gravity begins to pull it down
and we have an outside force.

The other questions seem pretty simple, for question 9,
All that matters is that the car is brought to a halt, within 0.13 s.
First calculate the accelaration, (18.5-0)/0.13s=
Then the force, F=mA.
Then for average force, f(ave)=F/t

The same formulae apply for questions 10 and 11