Short Answer
8.1 Linear Momentum, Force, and Impulse
If an object’s velocity is constant, what is its momentum proportional to?
 Its shape
 Its mass
 Its length
 Its breadth

Its impulse would be constant.

Its impulse would be zero.

Its impulse would be increasing.

Its impulse would be decreasing.

It is zero, because the net force is equal to the rate of change of the momentum.

It is zero, because the net force is equal to the product of the momentum and the time interval.

It is nonzero, because the net force is equal to the rate of change of the momentum.

It is nonzero, because the net force is equal to the product of the momentum and the time interval.
How can you express impulse in terms of mass and velocity when neither of those are constant?
 $\Delta \text{p}=\text{\Delta}(m\text{v})$
 $\frac{\text{\Delta}\text{p}}{\text{\Delta}t}=\frac{\text{\Delta}(m\text{v})}{\text{\Delta}t}$
 $\text{\Delta}\text{p}=\text{\Delta}(\frac{m}{\text{v}})$
 $\frac{\text{\Delta}\text{p}}{\text{\Delta}t}=\frac{1}{\text{\Delta}t}\text{\xb7}\text{\Delta}(m\text{v})$
How can you express impulse in terms of mass and initial and final velocities?
 $\text{\Delta}\text{p}\text{=}m\text{(}{\text{v}}_{\text{f}}{\text{v}}_{\text{i}}\text{)}$
 $\frac{\text{\Delta}\text{p}}{\text{\Delta}t}\text{=}\frac{m\text{(}{\text{v}}_{\text{f}}{\text{v}}_{\text{i}}\text{)}}{\text{\Delta}t}$
 $\text{\Delta}\text{p}\text{=}\frac{\text{(}{\text{v}}_{\text{f}}{\text{v}}_{\text{i}}\text{)}}{m}$
 $\frac{\text{\Delta}\text{p}}{\text{\Delta}t}=\frac{1}{m}\frac{({\text{v}}_{\text{f}}{\text{v}}_{\text{i}})}{\text{\Delta}t}$
Why do we use average force while solving momentum problems? How is net force related to the momentum of the object?
 Forces are usually constant over a period of time, and net force acting on the object is equal to the rate of change of the momentum.
 Forces are usually not constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval.
 Forces are usually constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval.
 Forces are usually not constant over a period of time, and net force acting on the object is equal to the rate of change of the momentum.
8.2 Conservation of Momentum
Under what condition(s) is the angular momentum of a system conserved?
 When net torque is zero
 When net torque is not zero
 When moment of inertia is constant
 When both moment of inertia and angular momentum are constant

It increases.

It decreases.

It stays constant.

It becomes zero.

Force is zero.

Force is not zero.

Force is constant.

Force is decreasing.
8.3 Elastic and Inelastic Collisions
Two objects collide with each other and come to a rest. How can you use the equation of conservation of momentum to describe this situation?
 m_{1}v_{1} + m_{2}v_{2} = 0
 m_{1}v_{1} − m_{2}v_{2} = 0
 m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}′
 m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{2}

Momentum is the sum of mass and velocity. Impulse is the change in momentum.

Momentum is the sum of mass and velocity. Impulse is the rate of change in momentum.

Momentum is the product of mass and velocity. Impulse is the change in momentum.

Momentum is the product of mass and velocity. Impulse is the rate of change in momentum.
What is the equation for conservation of momentum along the xaxis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest?
 m_{1}v_{1} = m_{1}v_{1}′cosθ_{1}
 m_{1}v_{1} = m_{1}v_{1}′cosθ_{1} + m_{2}v_{2}′cosθ_{2}
 m_{1}v_{1} = m_{1}v_{1}′cosθ_{1} − m_{2}v_{2}′cosθ_{2}
 m_{1}v_{1} = m_{1}v_{1}′sinθ_{1} + m_{2}v_{2}′sinθ_{2}
What is the equation for conservation of momentum along the yaxis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest?
 0 = m_{1}v_{1}′sinθ_{1}
 0 = m_{1}v_{1}′sinθ_{1} + m_{2}v_{2}′sinθ_{2}
 0 = m_{1}v_{1}′sinθ_{1} − m_{2}v_{2}′sinθ_{2}
 0 = m_{1}v_{1}′cosθ_{1} + m_{2}v_{2}′cosθ_{2}