Question

Suppose that we have two masses, m1 and m2, traveling at initial velocities v1i and v2i . After they collide, they will have velocities v1f and v2f . The collision will happen along a straight line, and there are no external forces involved.

Exercise 1 Let m1 = 0.20 kg, m2 = 0.50 kg, v1i = 0.50 m/s, and v2i = 0.0 m/s. If v2f = 0.286 m/s, what is v1f ?

For the collision above what are the initial and final kinetic energies? Is kinetic energy conserved in the collision?

Let m1 = 500 g, m2 = 500 g, v1i = 0.45 m/s, and v2i = 0.0 m/s. If v1f = 0.225 m/s, what is v2f ? Simply stated, what would you say about the two masses after the collision? What type of collision is this?

Answer 1

1.

By Conservation of momentum

M₁V_1i + M₂V_2i = M₁V_1f+ M₂V_2f

0.2*0.5+0.5*0 = 0.2*V_1f +0.5*0.286

V_1f = -0.215 m/s

Initial Kinetic energy is

K_initial = (1/2)M₁V_1i² +(1/2)M₂V_2i² =(1/2)(0.2)(0.5)² +(1/2)(0.5)(0)²=0.025 J

Final Kinetic energy is

K_final= (1/2)M₁V_1f² +(1/2)M₂V_2f² =(1/2)(0.2)(-0.215)² +(1/2)(0.5)(0.286)²=0.025 J

Since K_initial = K_final ,KInetic energy is conserved

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2.

By Conservation of momentum

M₁V_1i + M₂V_2i = M₁V_1f+ M₂V_2f

0.5*0.45+0.5*0 = 0.5*0.225+0.5*V_2f

V_2f = 0.225 m/s

After the collision the masses move with the same velocity in same direction.

Initial Kinetic energy is

K_initial = (1/2)M₁V_1i² +(1/2)M₂V_2i² =(1/2)(0.5)(0.45)² +(1/2)(0.5)(0)²=0.050625 J

Final Kinetic energy is

K_final= (1/2)M₁V_1f² +(1/2)M₂V_2f² =(1/2)(0.5)(0.225)² +(1/2)(0.5)(0.226)²=0.0253125 J

Since initial kinetic energy K_initial is not equal to final kinetic energy K_final.The Collision is Inelastic