In: Physics

A car of mass 500 kg is going towards east at 20.0 m/s. It hits a slow moving truck (moving at speed 10.0 m/s also going towards east) of mass 2000 kg from back. (a) If after the collision the truck is moving at 15.0 m/s towards east, what is the velocity of the car? (b) If the collision between the car and the truck is perfectly elastic, find their velocities after the collision. (c) Find the percentage loss of kinetic energy of the vehicles due to the collision in both cases.

**given**

**m1 = 500 kg
u1 = 20 m/s
m2 = 2000 kg
u2 = 10 m/s**

**a) v2 = 15 m/s
v1 = ?**

**Apply conservation of momentum**

**m1*u1 + m2*u2 = m1*v1 + m2*v2**

**v1 = (m1*u1 + m2*u2 - m2*v2)/m1**

**= (500*20 + 2000*10 - 2000*15)/500**

**= 0 m/s
<<<<<<<<<<<------------------Answer**

**b) if the collision perfectly elastic
collision,**

**after the collision,**

**velocity of the car,**

**v1 = ((m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)**

**= ( (500 - 2000)*20 + 2*2000*10)/(500 +
2000)**

**= 4 m/s
<<<<<<<<<<<------------------Answer**

**velocity of the truck,
v2 = ( (m2 - m1)*u2 + 2*m1*u1 )/(m1 + m2)**

**= ( (2000 - 500)*10 + 2*500*20 )/(500 +
2000)**

**= 14 m/s
<<<<<<<<<<<------------------Answer**

**c) in the first case, KEi = (1/2)*m1*u1^2 +
(1/2)*m2*u2^2**

**= (1/2)*500*20^2 + (1/2)*2000*10^2
= 200000 J**

**KEf = (1/2)*m1*v1^2 + (1/2)*m2*v2^2**

**= (1/2)*500*0^2 + (1/2)*2000*15^2
= 225000 J**

**here KEf > KEi, so kinetic energy is
gained.**

**percentage gained in kinetic energy = (KEf -
KEi)*100/KEi**

**= (225000 - 200000)*100/200000**

**= 12.5 %
<<<<<<<<<<<------------------Answer**

**in second case there is no gain or loss in kinetic
energy.**

**because, the collision is perfect elastic
collision.**

**so, in second case the percentage loss of kinetic energy
= 0 %
<<<<<<<<<<<------------------Answer**

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