Question

Block 1, of mass m1 = 3.90 kg , moves along a frictionless air track with speed v1 = 31.0 m/s . It collides with block 2, of mass m2 = 51.0 kg , which was initially at rest. The blocks stick together after the collision.
What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?

Answer 1

The energy would remain equal to that of the initial kinetic energy since it is a frictionless surface. However if it is an inelastic collision then energy would be lost due to sound and heat.

Answer 2

Answer:

ΔK= -1741. 09 J

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P=m*v

where:

p : Linear momentum

m : mass

v : velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀=Pf  Formula (1)

P₀ :Initial  linear momentum quantity

Pf : Initial  linear momentum quantity

Nomenclature and data

m₁:  block 1  mass = 3.90 kg

V₀₁: initial block 1 speed,  =31.0m/s

Vf₁: final  block 1  speed

m₂:  block 2 mass =  51.0 kg

V₀₂: initial  block 2  speed= 0

Vf₂: final  block 2  speed

Problem development

For this problem the collision is plastic ,then, the blocks stick together after the collision and Vf₁=Vf₂=Vf

We assume that the Block 1 moves to the right before the collision (+) and The joined blocks move to the right after the collision(+).

We apply formula (1)

P₀=Pf

m₁*V₀₁+m₂*V₀₂=m₁*Vf₁+m₂*Vf₂

m₁*V₀₁+m₂*V₀₂=(m₁+m₂) Vf

3.9*31+51*0=(3.9+51) Vf

120.9+0 = 54.9*Vf

Vf = 120.9/54.9

Vf = 2.2 m/s

ΔK in the two-block system's kinetic energy due to the collision

ΔK=Kfinal−Kinitial

ΔK: Change in kinetic energy (J)

Kfinal: final kinetic energy (J)

Kinitial : initial kinetic energy (J)

Kinitial=(1/2 )m₁*V₀₁²+(1/2 )m₂*V₀₂²= (1/2 )(3.9)*(31)²+(1/2 )(51)*0=1873.95 J

Kfinal = (1/2 )(m₁+m₂)*Vf²=  (1/2 )(3.9+51)* (2.2)² = 132.85 J

ΔK= 132.85 J−1873.95 J

ΔK= -1741. 09 J