Question

a 0.012kg rubber bullet travels at a velocity of 150m/s, hits a stationary 8.5kg concrete block resting on a frictionless surface, and ricochets in the opposite direction worth a velocity of -100m/s. how fast will the concrete block be moving?

Answer 1

Answer: The velocity of concrete block is 0.353 m/s

Step-by-step explanation:

To calculate the velocity of the concrete block after the collision, we use the equation of law of conservation of momentum, which is:

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

where,

`m_1` = mass of rubber bullet = 0.012 kg

`u_1` = Initial velocity of rubber bullet = 150 m/s

`v_1` = Final velocity of rubber bullet = -100 m/s

`m_2` = mass of concrete block = 8.5 kg

`u_2` = Initial velocity of concrete block = 0 m/s

`v_2` = Final velocity of concrete block = ?

Putting values in above equation, we get:

`(0.012* 150)+(8.5* 0)=(0.012* (-100))+(8.5* v_2)\\\\v_2=(1.8+1.2)/(8.5)=0.353m/s`

Hence, the velocity of concrete block is 0.353 m/s

Answer 2

This can be solve using momentum balance

Momentum = mv

Where m is the mass of the object

V is the velocity of the object

(0.012 kg) ( 150 m/s) + ( 8.5 kg) ( 0 m/s) = ( 0.012 kg) ( -100 m/s) + (v) ( 8.5 kg)

V = 0.35 m/s