Question

A 0.50kg ball traveling at 6.0 m/s collides head-on with a 1.00kg ball moving in the opposite direction at 12.0 m/s. After colliding, the 0.50kg ball bounces backward at 14 m/s. Find the other ball's speed and direction after the collision.

Answer 1

Answer:

Please check the 2nd page of the below file for the answer

Explanation:

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e46f78deb64a71c21a8d1b658769e1a8.pdf

Answer 2

Answer:

A 0.50kg ball traveling at 6.0 m/s collides head-on with a 1.00kg ball moving in the opposite direction at 12.0 m/s. After colliding, the 0.50kg ball bounces backward at 14 m/s. The other ball's speed will be 2 m/s and the direction after the collision will be the same negative direction.

Explanation:

We know,

Momentum (P) = Mass $\times$ velocity

Or,  P = mv

According to the conservation of linear momentum,

Momentum before collision =  Momentum after collision.

Sum of initial momentum two balls = Sum of final momentum of balls

Or,

$P_{1 i}+P_{2 i}=P_{1 f}+P_{2 f}$   ...(1)

That is,

$m_{1} v_{1 i}+m_{2} v_{2 i}=m_{1} v_{1 f}+m_{2} v_{2 f}$ ....(2)

Where,

$m_1$  - Mass of the first ball

$m_2$  - Mass of the second ball

$v_1{i}$  - Initial velocity of the first ball

$v_2_i$  - Initial velocity of the second ball

$v_1_f$ -  Final velocity of the first ball

$v_2_f$ - Final velocity of the second ball

From equation (2),

$\begin{array}{l}m_{2} v_{2 f}=m_{1} v_{1 i}+m_{2} v_{2 i}-m_{1} v_{1 f} \\v_{2 f}=\frac{m_{1} v_{1 i}+m_{2} v_{2 i}-m_{1} v_{1 f}}{m_{2}}\end{array}$

Let the original direction of the first ball be (+)

Given,

$m_1 = 0.50 \ kg$

$m_2 =1.00\ kg$

$v_1{i} = 6\ m/s$

$v_2_i = -12\ m/s$

Then,

$v_2_f$ $=\frac{0.50 \mathrm{~kg} \times 6.0 \frac{\mathrm{m}}{\mathrm{s}}+1.00 \mathrm{~kg} \times\left(-12.0 \frac{\mathrm{m}}{\mathrm{s}}\right)-0.50 \mathrm{~kg} \times\left(-14 \frac{\mathrm{m}}{\mathrm{s}}\right)}{1.00 \mathrm{~kg}}$

     = -2 m/s

The negative sign indicates that after the collision, the second ball will go in the same negative direction