Question

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.80 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use to mean to the right, - to mean to the left).A) What is the velocity of the first block after the collision?
B) What is the velocity of the second block after the collision?

Answer 1

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

$m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}$..........(1)

$m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s$

Substituting the above values into equation (1) and make V2 the subject of the formula:

$0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}$

$V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}$..................(2)

Using the law of conservation of kinetic energy:

$0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}$

$2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}$.......(3)

Substitute equation (2) into equation (3)

$2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s$

Substituting $V_{1}$ into equation(2)

$V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s$