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Question

Physics Question on work, energy and power

An object of mass 40kg40\, kg and having a velocity 4m/s4\, m/s collides with another object (m=60kg)\left( m=60\,kg \right) having velocity 2m/s2\, m/s. The collision is perfectly inelastic. The loss in energy is:

A

110 J

B

48 J

C

392 J

D

440 J

Answer

48 J

Explanation

Solution

When two bodies stick together after collision, then the collision is said to be perfectly inelastic.
In inelastic collision most of the kinetic energy is lost in other forms (specially as heat). Thus, in inelastic collision the kinetic energy is not conserved but total energy and momentum are conserved.
From law of conservation of momentum
m1u1+m2u2=(m1+m2)vm_{1} \,u_{1}+m_{2}\, u_{2}=\left(m_{1}+m_{2}\right) v
40×4+60×2=(40+60)v40 \times 4+60 \times 2=(40+60) v
v=280100=2.8m/s\Rightarrow v=\frac{280}{100}=2.8 \,m / s
Decrease in kinetic energy
=12m1v12+12m2v2212(m1+m2)v2=\frac{1}{2} m_{1}\, v_{1}^{2}+\frac{1}{2} m_{2}\, v_{2}^{2}-\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}
=12×40×(4)2+12×60×(2)212(40+60)(2.8)2=\frac{1}{2} \times 40 \times(4)^{2}+\frac{1}{2} \times 60 \times(2)^{2}-\frac{1}{2}(40+60)(2.8)^{2}
=320+120100×3.92=320+120-100 \times 3.92
=48J.=48 \,J .
Note: It is not necessary that in inelastic collision, there is always a loss of kinetic energy. If in collision, the potential energy of a body is released then the kinetic energy would increase.