Question

physics worksheet # 1: momentum = p = mv

1. calculate the momentum of a 1.60 x 10³ kg car traveling at 20.0 m/s.
2. calculate the momentum of a 2.50 x 10³ kg truck traveling at 110 km/h.
3. how fast is a 1.50 kg ball moving if it has a momentum of 4.50 kg.m/s?
4. a 75.0 g ball is rolling at a speed of 57.0 cm/s. calculate the balls momentum.
5. a 5.00 kg ball accelerates at a rate of 2.00 m/s² for 1.50 seconds. calculate the balls momentum after the acceleration.
6. a 2.00 kg rock is dropped from the top of a 30.0 m high building. calculate the balls momentum at the time that it strikes the ground.

Explanation:

1. Calculate the momentum of a \(1.60\times10^{3}\) kg car traveling at \(20.0\) m/s

Step1: Recall the momentum formula

Momentum formula is \(p = mv\), where \(m\) is mass and \(v\) is velocity.

Step2: Substitute the values

Given \(m=1.60\times 10^{3}\text{ kg}\) and \(v = 20.0\text{ m/s}\), then \(p=(1.60\times 10^{3})\times20.0\)

$$p = 3.20\times 10^{4}\text{ kg}\cdot\text{m/s}$$

2. Calculate the momentum of a \(2.50\times 10^{3}\) kg truck traveling at \(110\) km/h

Step1: Convert velocity to m/s

Since \(1\text{ km}=1000\text{ m}\) and \(1\text{ h}=3600\text{ s}\), \(v = 110\times\frac{1000}{3600}\text{ m/s}\approx30.56\text{ m/s}\)

Step2: Use the momentum formula

\(p=mv\), with \(m = 2.50\times 10^{3}\text{ kg}\) and \(v\approx30.56\text{ m/s}\), then \(p=(2.50\times 10^{3})\times30.56\)

$$p\approx7.64\times 10^{4}\text{ kg}\cdot\text{m/s}$$

3. How fast is a \(1.50\) kg ball moving if it has a momentum of \(4.50\) kg·m/s?

Step1: Rearrange the momentum formula for velocity

From \(p = mv\), we get \(v=\frac{p}{m}\)

Step2: Substitute the values

Given \(p = 4.50\text{ kg}\cdot\text{m/s}\) and \(m=1.50\text{ kg}\), then \(v=\frac{4.50}{1.50}\)

$$v = 3.00\text{ m/s}$$

4. A \(75.0\) g ball is rolling at a speed of \(57.0\) cm/s. Calculate the ball’s momentum.

Step1: Convert units

\(m=75.0\text{ g}=0.075\text{ kg}\) and \(v = 57.0\text{ cm/s}=0.57\text{ m/s}\)

Step2: Apply the momentum formula

\(p=mv=(0.075)\times(0.57)\)

$$p=0.04275\text{ kg}\cdot\text{m/s}$$

5. A \(5.00\) kg ball accelerates at a rate of \(2.00\) m/s² for \(1.50\) seconds. Calculate the ball’s momentum after the acceleration.

Step1: Find the final velocity using \(v = u+at\) (assuming initial velocity \(u = 0\))

\(v=0+(2.00)\times(1.50)=3.00\text{ m/s}\)

Step2: Calculate momentum

\(p=mv\), with \(m = 5.00\text{ kg}\) and \(v = 3.00\text{ m/s}\)

$$p=(5.00)\times(3.00)=15.0\text{ kg}\cdot\text{m/s}$$

6. A \(2.00\) kg rock is dropped from the top of a \(30.0\) m high building. Calculate the ball’s momentum at the time that it strikes the ground.

Step1: Use the kinematic equation \(v^{2}=u^{2}+2as\) ( \(u = 0\), \(a = g=9.8\text{ m/s}^{2}\), \(s = 30.0\text{ m}\))

\(v^{2}=0 + 2\times9.8\times30.0\), so \(v=\sqrt{588}\approx24.25\text{ m/s}\)

Step2: Calculate momentum

\(p=mv\), with \(m = 2.00\text{ kg}\) and \(v\approx24.25\text{ m/s}\)

$$p=(2.00)\times(24.25)=48.5\text{ kg}\cdot\text{m/s}$$

Answer:

1. \(3.20\times 10^{4}\text{ kg}\cdot\text{m/s}\)
2. \(\approx7.64\times 10^{4}\text{ kg}\cdot\text{m/s}\)
3. \(3.00\text{ m/s}\)
4. \(0.04275\text{ kg}\cdot\text{m/s}\)
5. \(15.0\text{ kg}\cdot\text{m/s}\)
6. \(48.5\text{ kg}\cdot\text{m/s}\)