Question

A body of mass 4kg moving with a velocity 12m/s collides with another mass of 6kg at rest. If two bidy stick together after collision then the loss of kinetic energy of system will be

Answer 1

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According to the law of conservation of linear momentum▪▪▪
m1v1=m2v2here m1=4,m2=6so 4×12=(4+6)×v2v2=4.8m/sLoss in kinetic energy is the difference in kinetic energy before and after collision=12×4×(12)2−12×10×(4.8)2=288−115.2=172.8J
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Answer 2

Answer:

The loss of kinetic energy of system is 172.8 J.

Explanation:

Given that,

Mass of object = 4 kg

Velocity of object = 12 m/s

Mass of another object = 6 kg

We need to calculate the velocity

Using conservation of momentum

$m_{1}v_{1}=(m_{1}+m_{2})v$

$v=\dfrac{m_{1}v_{1}}{m_{1}+m_{2}}$

Put the value in the formula

$v=\dfrac{4\times12}{10}$

$v=4.8\ m/s$

We need to calculate the loss of kinetic energy of system

Using formula of kinetic energy

$\Delta K.E=K.E_{i}-K.E_{f}$

$\Delta K.E=\dfrac{1}{2}m_{1}v_{1}^2-\dfrac{1}{2}mv^2$

Put the value into the formula

$\Delta K.E=\dfrac{1}{2}\times4\times12^2-\dfrac{1}{2}\times10\times(4.8)^2$

$\Delta K.E=172.8\ J$

Hence, The loss of kinetic energy of system is 172.8 J.