with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? O 0.50 Joules O 3.75 Joules 2.5 Joules O 1.25 Joules O 5 Joules

Principles of Physics: A Calculus-Based Text
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Chapter8: Momentum And Collisions
Section: Chapter Questions
Problem 1OQ
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A 10.0 kg block is sliding along a frictionless table at 1.0 m/s. It collides with a stationary block
which also has a mass of 10.0 kg. Immediately after the collision, the blocks stick together and move
with a speed of 0.5 m/s. How much kinetic energy is lost in the collision?
0.50 Joules
O 3.75 Joules
2.5 Joules
O 1.25 Joules
O 5 Joules
Transcribed Image Text:A 10.0 kg block is sliding along a frictionless table at 1.0 m/s. It collides with a stationary block which also has a mass of 10.0 kg. Immediately after the collision, the blocks stick together and move with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? 0.50 Joules O 3.75 Joules 2.5 Joules O 1.25 Joules O 5 Joules
Expert Solution
Step 1

Inelastic Collision

An inelastic collision is a type of collision where the kinetic energy is lost due to the internal friction of the particles. For a perfectly inelastic collision, the two colliding bodies stick together and move together as one unit. 

The above problem is an example of perfectly inelastic collision. Let two bodies of masses m1 and m2 be such that the first body is moving with an initial velocity v1i and the second body is at rest. 

The kinetic energy of the system before the collision

KEi=12m1v1i2+12m2.0=12m1v1i2

After the collision, the two bodies stick together and the two bodies move as a single unit. If vf is the velocity of the combined mass then the kinetic energy after the collision is

KEf=12m1+m2vf2

The loss of energy in this collision is

K=KEi-KEf=12m1v1i2-12m1+m2vf2

 

Step 2

Given in the question, the mass of the first block m1=10.0 kg

the mass of the second block m2=10.0 kg

 the initial velocity of the first block v1i=1 m/s.

the velocity of the combined system vf=0.5 m/s

 

Therefore the loss in kinetic energy

KE=12m1v1i2-12m1+m2vf2=12×10×12-12×10+10×0.52=2.5 J

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