Essential Physics

Section 1 review

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. Read the text aloud

Read the text aloud

momentum, impulse

\vec{p} = m \vec{v}

J = Δ p = F Δ t

Review problems and questions

State one similarity and one difference between inertia and momentum.

A hockey puck that has a mass of 170 g travels with a speed of 30 m/s.
1. What is the momentum of the puck?
2. 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: $\begin{array}{l} p & = & m v \\ = & (0.17 kg) \times (30 m / s) = 5.1 kg m / s \end{array}$ 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: $\begin{array}{l} J = p_{f} - p_{i} \\ = - 5.1 kg m/s - 5.1 kg m/s = - 10.2 kg m/s \end{array}$ 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.
1. What impulse was delivered to the ball?
2. The ball and Julieta’s foot experience equal-but-opposite 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: $\begin{array}{l} p_{i} = m v_{i} = 0 \\ p_{f} = m v_{f} = (0.2 kg) (20 m/s) = 4 kg m/s \\ p_{f} - p_{i} = 4 kg m/s - 0 = 4 kg m/s \\ J = p_{f} - p_{i} = 4 kg m/s \end{array}$ 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: $\begin{matrix} J = F Δ t & \Rightarrow & Δ t = \frac{J}{F} = \frac{4 kg m/s}{40 N} = 0.1 s \end{matrix}$ 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} Marco: J = (80 N) (0.05 s) = 4 N s \\ Marcela: J = (60 N) (0.1 s) = 6 N s \end{array}

Although Marco strikes his bag more forcefully, Marcela delivers a larger impulse. (Note that the unit newton-second, 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} = (0.200 kg) (- 10.0 m/s) = - 2.0 kg m/s

The final momentum of the ball will be positive:

p_{f} = m v_{f} = (0.200 kg) (+ 6.0 m/s) = + 1.2 kg m/s

The change in momentum Δp is therefore

Δ p = p_{f} - p_{i} = (+ 1.2 kg m/s) - (- 2.0 kg m/s) = + 3.2 kg m/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?

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_{e l e p h a n t} = m v = (6000 kg) (11 m/s) = 66, 000 kg m/s \\ p_{b u l l e t} = m v = (0.200 kg) (300 m/s) = 60 kg m/s \\ p_{e n g i n e} = m v = (18, 000 kg) (0 m/s) = 0 kg m/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 = Δ p = F Δ t \\ F = \frac{Δ p}{Δ t} = \frac{(0 kg m/s) - (65 kg) (+ 8.0 m/s)}{4.0 s} = \frac{- 520 kg m/s}{4.0 s} = - 130 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 = m v = (1, 000 kg) (- 40 m/s) = - 40, 000 kg m/s


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