Question

Two carts are colliding on an airtrack (neglect friction). The first cart has a mass of m1 = 40 g and an initial velocity of v1 = 2 m/s. The second cart has a mass of m2 = 47 g and an initial velocity of v2 = -5 m/s. Two experiments are conducted. In the first experiment, the first cart has a final velocity of v1' = -1.11 m/s. What is the velocity of the second cart?For the second experiment, the bumpers of the carts are modified, but the carts are started with the same initial velocities as before. Now the first cart has a final velocity of v1' = -3.8942 m/s, and the second cart a final velocity of v2' = 0.0163 m/s. How much (if any) energy was lost in the collision?In good approximation, what kind of collision was the second experiment?a. Totally inelasticb. Elasticc. Inelastic, but not totally

Answer 1

a) v2 = -2.5319m/s

b) losses = 0.3642 J

Step-by-step explanation:

Data Given:

`m_(1) = 0.04 kg\\v_(1) = +2m/s , v'_(1) = -1.11m/s \\\\m_(2) = 0.047 kg\\v_(2) = -5m/s\\\\`

part a

Using conservation of momentum:

`P_(f) = P_(i)\\m_(1)*v_(1) + m_(2)*v_(2) = m_(1)*v'_(1) + m_(2)*v'_(2) \\\\v'_(2) = (m_(1)*v_(1) + m_(2)*v_(2) - m_(1)*v'_(1) )/(m_(2)) \\\\v'_(2) = (0.04*2 + 0.047*-5 - 0.04*-1.11 )/(0.047) \\\\v'_(2) = -2.5319 m/s`

part b

Using energy conservation:

`E_(1) + E_(2) = E'_(1) + E'_(2) + Losses\\\\Losses = E_(1) - E'_(1) + E_(2) - E'_(2)\\\\Losses = 0.5*m_(1)*(v^2_(1) - v' ^2_(1)) + 0.5*m_(2)*(v^2_(2) - v' ^2_(2)) \\\\Losses = 0.5*(0.04)*((2)^2 - (-3.8942)^2) + 0.5*(0.047)*((-5)^2 - (0.0163)^2)\\\\Losses = 0.3642 J`

Answer 2

v₂ = -2.35 m/s ΔK = -0.364 J c)

Step-by-step explanation:

a) For the first experiment, assuming no external forces acting during the collision, total momentum must be conserved.

⇒ Δp = 0 ⇒ p₀ = pf

p₀ = m₁*v₀₁ + m₂*v₀₂ = (.04 kg)* (2 m/s) + (.047 kg)* (-5 m/s) =-0.155 kg*m/s

pf = m₁*vf₁ + m₂*vf₂ = (0.04 kg)* (-1.11 m/s) + (0.047 kg)*(vf₂) = -0.155 kg*m/s

Solving for vf₂:

vf₂ = (-0.155 kg*m/s + 0.044 kg*m/s) / 0.047 kg = -2.35 m/s

b) We need to get first the initial kinetic energy of both masses, as follows:

K₀ = K₀₁ + K₀₂ = 1/2*(0.04 kg)(2m/s)² + 1/2*(0.047 Kg)*(-5m/s)² = 0.67 J

The final kinetic energy will be the sum of the final kinetic energies of both masses, as follows:

Kf = Kf₁ + Kf₂ = 1/2*(0.04 kg)(-3.8942 m/s)² + 1/2*(0.047 kg)*(0.0163 m/s)²

⇒ Kf = 0.303 J

⇒ ΔK = 0.303 J - 0.67 J = -0.364 J

In terms of loss of kinetic energy, the collision can be classified as inelastic, even though not completely, as both masses didn't stick together.