Solution

A. Problem 1:

What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s?

Simple multiplication using the definition of momentum:

P = mb = (0.028 kg)(8.4 m/s) = 0.24 kg m/s

B. Problem 2:

A constant friction force of 25 N acts on a 65-kg skier for 20 s. What is the skier’s change in velocity?

Use Newton's second law in momentum form and solve for delta v

Assume the skier is moving to the right and let this be positive. The friction then acts in a netative direction

F = ma gives FD t = m D v gives D v = F D t/m

Using the values given, we have D v = -(25 N)(20 s)/65 kg = -7.7 m/s

C. Problem 3:

A 0.145-kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s.

If the contact time between bat and ball is 3.00 x 10^-3 s, calculate the average force between the ball and bat during contact.

Same as above except solve for average force as mxdelta v/delta t

We can let the positive direction be from batter toward pitcher

F = m D v/D t = (0.145 kg)(39 m/s - (-52 m/s))/3.0x10^-3 s = 4.40 x 10-3 N, towards the pitcher (positive direction)

D. Problem 5:

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of 4.0 x 10⁴

m/s (at the moment of takeoff).

The force on the gas can be found from its change in momentum.

The force on the rocket is the Newton’s 3^rd law pair (equal and opposite) to the force on the gas, and so is the opposite of the above.

FD t = m D v

If we can find the force on the gas, then, from Newton's 3rd law, the force on the rocket is equal and opposite to this force

The force on the gas is its change in momentum

F = vD m/Dt = (4.0 x 10⁴ m/s)(1500 kg/s) = 6.0 x 10⁷ N downward

So, by the third law, the force on the rocket is 6.0 x 10⁷ N upward