Question

A rubber block (m= 20.0g) slides along a frictionless table at v=2.00m/s directly towards a steel block (m= 0.0500kg), which is at rest on the table. After an inelastic collision, in which 25% of the system's kinetic energy is lost, the rubber block rebounds the way it came at a reduced speed. Determine the speed of each block after the collision.

(Is there a way to solve without quadratic?)

Answer 1

Using momentum conservation

Pi = Pf

m1V1i + m2V2i = m1V1f + m2*V2f

given that m1 = mass of rubber block = 20.0 g = 0.0200 kg

& m2 = mass of steel block = 0.0500 kg

V1i = initial speed of m1 = 2.00 m/s

V2i = initial speed of m2 = 0 m/s

initial speed of m1 = 2.00 m/s

V1f and V2f = ??

So, 0.0200*2.00 + 0.0500*0 = 0.02*V1f + 0.05*V2f

4 = 2*V1f + 5*V2f eq (1)

Now given 25% loss in System kinetic energy in collision.

So,KEf = 0.75*KEi

(0.5*m1*V1f^2) + (0.5*m2*V2f^2) = 0.75*((0.5*m1*V1f^2) + (0.5*m2*V2f^2))

using known values,

(0.5*0.02*V1f^2) + (0.5*0.05*V2f^2) = 0.75*0.5*0.02*2.0^2 = 0.03

6 = 2*V1f^2 + 5*V2f^2 eq(2)

Solving both eq(1) and (2),

V1f = -0.58 m/s and V2f = 1.03 m/s

Or V1f = 1.72 m/s and V2f = 0.11 m/s

GIven, the rubber block rebounds the way it came at a reduced speed.

So,

V1f = speed of rubber block = -0.58 m/s

V2f = speed of steel block = +1.03 m/s

No, there is now way to solve without Quadratic.

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