Question

A ball of 4 kg mass moving with velocity 40 m/s in
east direction collides with another ball of 6 kg
mass moving with velocity 20 m/s in east direction.
After collision two balls stick together, then loss of
kinetic energy is​

Answer 1

Answer:

480 joules

Explanation:

m1u1+m2u2=(M1+m2)v

4×40+6×20=(6+4)v

160+120=10v

v=280/10=28m/s

K.E before collision

=1/2×mu1²+1/2×mu2²

=1/2×4×40×40+1/2×6×20×20

=4400Joules

K.E after collision

1/2×(M1+m2)×v²

=1/2×10×28×28

=3920 joules

Loss of K.E=4400J-3920J

=480joules

Answer 2

Answer:

Loss in Kinetic Energy is $480J$.

Explanation:

This is based on the conservation of momentum.

It states that the momentum before and after the collision is constant.

It remains conserved.

It is given that a ball of $4 kg$ mass moving with velocity $40 m/s$ in

east direction collides with another ball of $6 kg$ mass moving with velocity $20 m/s$ in east direction.

We need to determine loss of kinetic energy.

According to law of conservation of momentum,

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

Using values

$4*40+6*20=(6+4)v$

$v=28m/s$

Kinetic energy before collision,

$\frac{1}{2} m_{1} u_{1}^{2} +\frac{1}{2}m_{2} u_{2}^{2}$

$=4400J$

KE after collision,

$\frac{1}{2} ( m_{1}+m_{2})v^{2}$

$=3920J$

Loss is KE is $4400-3920=480J$

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