Question

1. give an example of a system whose mass is not constant?

a. a spinning top
b. a baseball flying through the air
c. a rocket launched from earth
d. a block sliding on a frictionless inclined plane

1. a soccer ball with a mass of 0.4 kg is kicked with a force on 50 n for 0.1 seconds. what is the change in momentum of the ball?
2. a truck with a mass of 2000 kg is traveling at 15 m/s. the driver applies the brakes, exerting a force of 4000 n for 10 seconds. what is the trucks final speed?

Explanation:

Question 6

A rocket launched from Earth burns fuel, which is expelled as exhaust. This means the mass of the rocket (system) decreases over time. A spinning top, a baseball flying through the air, and a block on a frictionless incline have constant mass (assuming no parts are lost or added).

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that the change in momentum \(\Delta p\) of an object is equal to the impulse \(J\) applied to it. The formula for impulse is \(J = F\times\Delta t\), where \(F\) is the force and \(\Delta t\) is the time interval.

Step2: Substitute the given values

Given \(F = 50\space N\) and \(\Delta t=0.1\space s\). Using the formula \(J=\Delta p=F\times\Delta t\), we substitute the values: \(\Delta p=(50\space N)\times(0.1\space s)\)

Step3: Calculate the change in momentum

\(\Delta p = 5\space kg\cdot m/s\)

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem is \(J=\Delta p\), where \(J = F\times\Delta t\) and \(\Delta p=m(v - u)\). Here, \(u = 15\space m/s\), \(m = 2000\space kg\), \(F=- 4000\space N\) (negative because it's a braking force), and \(\Delta t = 10\space s\)

Step2: Substitute into the impulse - momentum formula

\(F\times\Delta t=m(v - u)\)
\(-4000\times10=2000\times(v - 15)\)

Step3: Solve for \(v\)

First, simplify the left - hand side: \(-40000 = 2000v-30000\)
Then, add \(30000\) to both sides: \(-40000 + 30000=2000v\)
\(-10000 = 2000v\)
Divide both sides by \(2000\): \(v=- 5\space m/s\). But speed cannot be negative in the context of magnitude (assuming we consider the magnitude of velocity as speed here, and the negative sign just indicates direction opposite to the initial motion). So the final speed is \(5\space m/s\)

Answer:

C. A rocket launched from Earth

Question 7