
Momentum is a vector quantity: the product of mass and velocity. Impulse (also a vector) is the change in an object’s momentum. Impulse can be expressed as the product of an applied force and the force’s duration. Cushioning increases the time interval for momentum transfer, and this reduces the forces felt in car crashes and contact sports.

momentum, impulse


$$\overrightarrow{p}=m\overrightarrow{v}$$


$$J=\text{\Delta}\text{\hspace{0.05em}}p=F\text{\Delta}t$$



Review problems and questions 

 State one similarity and one difference between inertia and momentum.

Inertia and momentum both increase with an object’s mass, and both concepts indicate how easy or hard it will be to change the object’s motion. However, momentum only exists when an object is in motion, whereas inertia describes an object even if it is stationary.


A hockey puck that has a mass of 170 g travels with a speed of 30 m/s.
 What is the momentum of the puck?
 What impulse must be imparted on the puck by a player who wishes to change the puck’s direction by 180°, while keeping the puck moving at the same speed?

Answer:  The momentum of the puck is 5.1 kg m/s.
 The impulse imparted on the puck is −10.2 kg m/s, which is twice the magnitude of the initial momentum, but opposite in sign.
Solution:  Asked: momentum p of the puck
Given: puck mass m = 170 g = 0.17 kg, velocity v = 30 m/s Relationships: p = mv Solve: $$\begin{array}{lll}p\hfill & =\hfill & mv\hfill \\ \hfill & =\hfill & \left(0.17\text{\hspace{0.17em}}\text{kg}\right)\times \left(30\text{\hspace{0.17em}}\text{m}/\text{s}\right)=5.1\text{\hspace{0.17em}}\text{kg}\text{\hspace{0.17em}}\text{m}/\text{s}\hfill \end{array}$$Answer: The momentum of the puck is 5.1 kg m/s.  Asked: impulse J imparted by the hockey player
Givens: puck initial momentum p_{i} = +5.1 kg m/s, final momentum p_{f} = −5.1 kg m/s Relationships: J = Δp = p_{f} − p_{i} Solve:$$\begin{array}{l}J={p}_{f}{p}_{i}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5.1\text{kg m/s}5.1\text{kg m/s}=10.2\text{kg m/s}\end{array}$$Answer: The impulse imparted on the puck is −10.2 kg m/s, which is twice the magnitude of the initial momentum, but opposite in sign.

 Julieta kicks a stationary 0.2 kg soccer ball. The ball leaves her foot at a speed of 20 m/s.
 What impulse was delivered to the ball?
 The ball and Julieta’s foot experience equalbutopposite forces, each 40 N strong. How long (in seconds) were the ball and foot in contact?

Answer:  Julieta’s foot delivered an impulse of 4 kg m/s to the ball.
 The ball and foot were in contact for 0.1 s.
Solution:  Asked: impulse J delivered to the ball
Givens: ball mass m = 0.2 kg, initial velocity v_{i} = 0, final velocity v_{f} = 20 m/s Relationships: p = mv; J = p_{f} − p_{i} Solve:$$\begin{array}{l}{p}_{i}=m{v}_{i}=0\\ {p}_{f}=m{v}_{f}=(0.2\text{kg})(20\text{m/s})=4\text{kgm/s}\\ {p}_{f}{p}_{i}=4\text{kgm/s}0=4\text{kgm/s}\\ J={p}_{f}{p}_{i}=4\text{kgm/s}\end{array}$$Answer: Julieta’s foot delivered an impulse of 4 kg m/s to the ball.  Asked: time duration of contact Δt
Givens: impulse J = 4.0 kg m/s, force F = 40 N Relationship: J = FΔt Solve:$$\begin{array}{ccc}J=F\Delta t& \Rightarrow & \Delta t=\frac{J}{F}=\frac{4\text{kgm/s}}{40\text{N}}=0.1\text{s}\end{array}$$Answer: The ball and foot were in contact for 0.1 s.

 Marco strikes a punching bag with 80 N of force for 0.05 s. Marcela strikes a body bag with 60 N of force for 0.1 s. Who delivers a greater amount of impulse?

Answer: Marcela delivers a larger impulse.
Asked: impulse J delivered to the bag Givens: force (Marco) F = 80 N; time interval (Marco) Δt = 0.05 s; force (Marcela) F = 60 N; time interval (Marcela) Δt = 0.1 s Relationships: J = FΔt Solve: $$\begin{array}{l}\text{Marco:}J=(80\text{N})(0.05\text{s})=4\text{Ns}\\ \text{Marcela:}J=(60\text{N})(0.1\text{s})=6\text{Ns}\end{array}$$Although Marco strikes his bag more forcefully, Marcela delivers a larger impulse. (Note that the unit newtonsecond, or N s, is equivalent to kilogram meter per second, or kg m/s. This means that the impulse delivered by each kickboxer equals the change in momentum of their respective targets.)

 A ball with a mass of 200 g is thrown straight down at the floor. It strikes the floor at a speed of 10.0 m/s and bounces straight up again with a speed of 6.0 m/s.
What is the change in the ball’s momentum?

Answer: The change in momentum of the ball is 3.2 kg m/s.
Solution: Let down be the negative direction. The initial momentum of the ball will then be negative: $${p}_{i}=m{v}_{i}=\left(0.200\text{kg}\right)\left(10.0\text{m/s}\right)=2.0\text{kgm/s}$$The final momentum of the ball will be positive: $${p}_{f}=m{v}_{f}=\left(0.200\text{kg}\right)\left(+6.0\text{m/s}\right)=+1.2\text{kgm/s}$$The change in momentum Δp is therefore $$\text{\Delta}p\text{=}{p}_{f}{p}_{i}=\left(+1.2\text{kgm/s}\right)\left(2.0\text{kgm/s}\right)=+3.2\text{kgm/s}$$

 A 1,200 kg car is heading due north at 14.0 m/s. It collides with an identical car heading due south at 14.0 m/s. What is the combined momentum of the cars before the collision?

Answer: The combined momentum of the cars before the collision is zero, or 0 kg m/s.
Solution: Let north be the positive direction. The car heading north has momentum p, and the car heading south with the same mass and same speed has momentum −p. Adding these momenta together gives a total momentum of zero.

 Which of the following has greatest inertia? Which has greatest momentum?
 6,000 kg elephant charging at 11 m/s
 200 g bullet fired at 300 m/s
 18,000 kg fire engine parked on the street

Answer: The fire engine has the greatest mass and therefore the greatest inertia, but the elephant has the greatest momentum.
Solution: $$\begin{array}{l}{p}_{elephant}=mv=\left(6000\text{kg}\right)\left(11\text{m/s}\right)=66,000\text{kgm/s}\\ {p}_{bullet}=mv=\left(0.200\text{kg}\right)\left(300\text{m/s}\right)=60\text{kgm/s}\\ {p}_{engine}=mv=\left(18,000\text{kg}\right)\left(0\text{m/s}\right)=0\text{kgm/s}\end{array}$$

 Sean, whose mass is 60 kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He brakes the sled with a constant force, bringing it to a stop in 4.0 s. What force does he apply?

Answer: The force applied is −130 N.
Solution:$$\begin{array}{l}J=\text{\Delta}p\text{=}F\text{\Delta}t\text{}\\ F\text{=}\frac{\text{\Delta}p}{\text{\Delta}t\text{}}\text{=}\frac{\left(0\text{kgm/s}\right)\left(65\text{kg}\right)\left(+8.0\text{m/s}\right)}{4.0\text{s}}=\text{}\frac{520\text{kgm/s}}{4.0\text{s}}=130\text{N}\end{array}$$

 A 1000 kg car is traveling north on a highway at 20 m/s.
What is the car’s momentum when observed by another car traveling in the opposite direction at 20 m/s?

Answer: −40,000 kg m/s.
Solution: The momentum depends on the reference frame of the observer—in this case, the car traveling in the opposite direction. In this reference frame, the first car is traveling at (20 + 20) m/s = 40 m/s in the opposite direction, or −40 m/s in this reference frame. The momentum of the first car is therefore$$p=mv=\left(1,000\text{kg}\right)\left(40\text{m/s}\right)=40,000\text{kgm/s}$$

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