Question

A body of mass 40kg having velocity 4 m/s collides with another body of mass 60kg having velocity 2 m/s. If the collision is inelastic, then loss in kinetic energy will be

Answer 1

m1u1+m2u2=m1v1+m2v2
40(4)+60(2)=40(x)+60(0)
160+120=40x
x=280/40=7m/s

Answer 2

The loss in kinetic energy will be 48 J.

Explanation:

It is given that,

Mass of body 1, $m_1=40\ kg$

Initial speed of body 1, $u_1=4\ m/s$

Mass of body 2, $m_2=60\ kg$

Initial speed of body 2, $u_2=2\ m/s$

To find,

The loss in kinetic energy.

Solution,

In case of inelastic collision, two objects move with a common velocity. The momentum remains conserved such that,

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

$V=\dfrac{40\times 4+60\times 2}{40+60}$

V = 2.8 m/s

Initial kinetic energy of the system,

$K_i=\dfrac{1}{2}(m_1u_1^2+m_2u_2^2)$

$K_i=\dfrac{1}{2}(40(4)^2+60(2)^2)$

$K_i=440\ J$

Final kinetic energy of the system is :

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}\times (40+60)\times (2.8)^2$

$K_f=392\ J$

The loss in kinetic energy will be :

$\Delta K=K_f-K_i$

$\Delta K=392-440$

$\Delta K=-48\ J$

So, the loss in kinetic energy will be 48 J.

Learn more,

Inelastic collision

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