Question

1. a 0.530 - kg basketball hits a wall head - on with a forward speed of 18.0 m/s. it rebounds with a speed of 13.5 m/s. the contact time is 0.100 seconds.

a. determine the impulse with the wall
b. determine the force of the wall on the ball

1. a 40.0 g hot dog moving with a velocity of 9.00 m/s to the south collides with a 55.0 g hot dog bun with a velocity of 6.00 m/s to the north. if the two objects stick together upon collision, what is the velocity of the combined masses?
2. a 1200 kg car travelling at 20.0 m/s speeds up to 30.0 m/s. what is the impulse experienced by the car?

Explanation:

Step1: Calculate impulse for problem 8a

Impulse \(J=\Delta p = m(v_f - v_i)\). Let the initial velocity \(v_i = 18.0\ m/s\) (forward) and final velocity \(v_f=- 13.5\ m/s\) (rebound). Mass \(m = 0.530\ kg\).

$$J=0.530\times(-13.5 - 18.0)$$

$$J=0.530\times(-31.5)$$

$$J=- 16.7\ kg\cdot m/s$$

(The negative sign indicates direction opposite to the initial velocity)

Step2: Calculate force for problem 8b

Impulse - force relation \(J = F\Delta t\). We know \(J=-16.7\ kg\cdot m/s\) and \(\Delta t = 0.100\ s\)

$$F=\frac{J}{\Delta t}=\frac{-16.7}{0.100}=-167\ N$$

(The negative sign indicates the force is in the direction opposite to the initial velocity of the ball)

Step3: Solve problem 9

Use conservation of momentum \(m_1v_1 + m_2v_2=(m_1 + m_2)v\). \(m_1 = 40.0\ g=0.0400\ kg\), \(v_1=-9.00\ m/s\) (south), \(m_2 = 55.0\ g = 0.0550\ kg\), \(v_2 = 6.00\ m/s\) (north)

$$0.0400\times(-9.00)+0.0550\times6.00=(0.0400 + 0.0550)v$$

$$ - 0.36+0.33=0.095v$$

$$v=\frac{-0.03}{0.095}\approx - 0.316\ m/s$$

(The negative sign indicates the combined mass moves to the south)

Step4: Solve problem 10

Impulse \(J=\Delta p=m(v_f - v_i)\). \(m = 1200\ kg\), \(v_i = 20.0\ m/s\), \(v_f = 30.0\ m/s\)

$$J=1200\times(30.0 - 20.0)$$

$$J=1200\times10=12000\ kg\cdot m/s$$

Answer:

a. The impulse is \(-16.7\ kg\cdot m/s\)
b. The force is \(-167\ N\)

1. The velocity is \(-0.316\ m/s\)
2. The impulse is \(12000\ kg\cdot m/s\)