Question

question 14 of 20
a 0.6 kg basketball moving 7.2 m/s to the right collides with a 0.04 kg tennis ball moving 22 m/s to the left. after the collision, the tennis ball is moving 32.7 m/s to the right. what is the velocity of the basketball after the collision? assume an elastic collision occurred.
a. 3.6 m/s to the left
b. 10.9 m/s to the right
c. 3.6 m/s to the right
d. 10.9 m/s to the left

Explanation:

Step1: Define variables and initial - final directions

Let the right - hand direction be positive. $m_1 = 0.6$ kg (mass of basketball), $v_{1i}=7.2$ m/s (initial velocity of basketball), $m_2 = 0.04$ kg (mass of tennis ball), $v_{2i}=- 22$ m/s (initial velocity of tennis ball), $v_{2f}=32.7$ m/s (final velocity of tennis ball), and $v_{1f}$ is the final velocity of the basketball we want to find.

Step2: Apply the law of conservation of momentum

The law of conservation of momentum is $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$.
Substitute the known values:

$$
LATEXBLOCK0
$$

Step3: Solve for $v_{1f}$

First, subtract 1.308 from both sides:

$$
LATEXBLOCK1
$$

Then, divide both sides by 0.6:

$$v_{1f}=\frac{2.132}{0.6}=3.553\approx3.6\text{ m/s}$$

Since $v_{1f}>0$, the basketball is moving to the right.

Answer:

C. 3.6 m/s to the right