Question

A 0.80 kg basketball traveling upward at 5.0 m/s impacts an 8.0 10 kg tennis ball traveling downward at 5.0 m/s. The basketball’s velocity after the collision is 3.0 m/s upward. What is the velocity of the tennis ball after the collision?

Answer 1

To solve this problem we will apply the concepts related to the conservation of momentum. This can be defined as the product between the mass and the velocity of each object, and by conservation it will be understood that the amount of the initial momentum is equal to the amount of the final momentum. By the law of conservation of momentum,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_1$ = Mass of Basketball

$m_2$ = Mass of Tennis ball

$u_1$ = Initial velocity of Basketball

$u_2$ = Initial Velocity of Tennis ball

$v_1$ = Final velocity of Basketball

$v_2$ = Final velocity of the tennis ball

Replacing,

$(0.8)(0.5)+(0.1)(-5.0)=(0.8)(0.3)+(0.1)v_2$

Solving for the final velocity of the tennis ball

$v_2 = 11m/s^2$

Therefore the velocity of the tennis ball after collision is 11 m/s

Answer 2

Corrected Question:

A 0.80 kg basketball traveling upward at 5.0 m/s impacts an 8.0 x $10^{-2}$ kg tennis ball traveling  downward at 5.0 m/s. The basketball’s velocity after the collision is 3.0 m/s upward. What is the  velocity of the tennis ball after the collision?

Source: http://physics.ggould.com/downloads/momentum_MC+problems.pdf

Answer:

15m/s

Explanation:

The principle of conservation of linear momentum states that "in the absence of external forces, the linear momentum of colliding bodies before and after collision does not change".

This means that total linear momentum of colliding bodies before collision is the same after collision.

The momentum (P) of a body is the motion that the body has and it is given by the product of the mass (m) of the body and its velocity (v). i.e

P = m x v

Using this principle, lets solve the problem;

Momentum of basket ball before collision + Momentum of tennis ball before collision

    =

Momentum of basket ball after collision + Momentum of tennis ball after collision.

Let;

upward motion be positive and downward be negative.

Also let,

$m_{B}$ = mass of basketball = 0.80kg

$m_{T}$ = mass of tennis ball = 8 x $10^{-2}$kg = 0.08kg

$u_{B}$ = initial velocity of the basket ball before collision = +5.0m/s (since motion is upwards).

$u_{T}$ = initial velocity of the tennis ball before collision = -5.0m/s (since motion is downwards).

$v_{B}$ = final velocity of the basketball after collision = +3.0m/s (since motion is upwards).

$v_{T}$ = final velocity of the tennis ball after collision.

Therefore,

Momentum of basket ball before collision = $m_{B}$ x $u_{B}$

Momentum of tennis ball before collision = $m_{T}$ x $u_{T}$

Momentum of basket ball after collision = $m_{B}$ x $v_{B}$

Momentum of tennis ball after collision = $m_{T}$ x $v_{T}$

=> ($m_{B}$ x $u_{B}$) + ($m_{T}$ x $u_{T}$) = ($m_{B}$ x $v_{B}$) + ($m_{T}$ x $v_{T}$)

Substituting the values into the above equation gives;

=> (0.8 x 5.0) + (0.08 x -5.0) = (0.8 x 3.0) + (0.08 x $v_{T}$)

=> (0.8 x 5.0) - (0.08 x 5.0) = (0.8 x 3.0) + (0.08 x $v_{T}$)

=> (4) - (0.4) = (2.4) + (0.08$v_{T}$)

=> 3.6 = 2.4 + 0.08$v_{T}$

=> 0.08$v_{T}$ = 3.6 - 2.4

=> 0.08$v_{T}$ = 1.2

=> $v_{T}$ = $\frac{1.2}{0.08}$

=> $v_{T}$ = 15m/s

Therefore, the velocity of the tennis ball after the collision is 15m/s