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?

Answers

v₂ = -2.35 m/s ΔK = -0.364 J c)

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.

a) v2 = -2.5319m/s

b) losses = 0.3642 J

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} = \frac{m_{1}*v_{1} + m_{2}*v_{2} - m_{1}*v'_{1} }{m_{2}} \\\\v'_{2} = \frac{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

v₂ = -2.35m/s

This is an Inelastic collision

Explanation:

Law of conservation of momentum

This states that for a collision occurring between two object in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

Given from the question

m₁=40 g

initial velocity of m₁: v₁=2 m/s

m₂=47 g

initial velocity of m₂: v₂ = -5 m/s

final velocity of m₁: v'₁ = -1.11 m/s.

final velocity of  m₂ = ?

second experiment final velocity of m₁ : v₁' = -3.8942 m/s

second experiment final velocity of m₂ : v₂' = 0.0163 m/s

Considering the first experiment

Apply the knowledge of conservation of momentum

m₁v₁ + m₂v₂ =m₁v₁ + m₂v₂

(0.040)(2) + (0.047)(-5) = (0.040)(-1.11) + (0.040)v₂

v₂ = -2.35m/s

Considering the second experiment

initial kinetic energy KEₓ  = (1/2)m₁v₁² + (1/2)m₂v₂²

     KEₓ = (1/2)(0.040)(2)² + (1/2)(0.047)(-5)²

     KEₓ = 0.6675 J

Final kinetic energyKEₙ = (1/2)m₁v₁² + (1/2)m₂v₂²


KEₙ = (1/2) (0.040) (-3.8942)² + (1/2) (0.047) (0.0163)²

KEₙ = 0.3033 J

Loss in kinetic energy  ΔKE = KEₓ - KEₙ

ΔKE = 0.6675 - 0.3033

ΔKE = 0.3642 J

This collision is a perfectly inelastic collision because the maximum  kinetic energy is lost this means that the kinetic energy before the collision is not the same as the kinetic energy after the collision.Though momentum is conserved kinetic energy is not conserved.

v₂ = -2.35m/s

This is an Inelastic collision

Explanation:

Law of conservation of momentum

This states that for a collision occurring between two object in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

Given from the question

m₁=40 g

initial velocity of m₁: v₁=2 m/s

m₂=47 g

initial velocity of m₂: v₂ = -5 m/s

final velocity of m₁: v'₁ = -1.11 m/s.

final velocity of  m₂ = ?

second experiment final velocity of m₁ : v₁' = -3.8942 m/s

second experiment final velocity of m₂ : v₂' = 0.0163 m/s

Considering the first experiment

Apply the knowledge of conservation of momentum

m₁v₁ + m₂v₂ =m₁v₁ + m₂v₂

(0.040)(2) + (0.047)(-5) = (0.040)(-1.11) + (0.040)v₂

v₂ = -2.35m/s

Considering the second experiment

initial kinetic energy KEₓ  = (1/2)m₁v₁² + (1/2)m₂v₂²

     KEₓ = (1/2)(0.040)(2)² + (1/2)(0.047)(-5)²

     KEₓ = 0.6675 J

Final kinetic energyKEₙ = (1/2)m₁v₁² + (1/2)m₂v₂²


KEₙ = (1/2) (0.040) (-3.8942)² + (1/2) (0.047) (0.0163)²

KEₙ = 0.3033 J

Loss in kinetic energy  ΔKE = KEₓ - KEₙ

ΔKE = 0.6675 - 0.3033

ΔKE = 0.3642 J

This collision is a perfectly inelastic collision because the maximum  kinetic energy is lost this means that the kinetic energy before the collision is not the same as the kinetic energy after the collision.Though momentum is conserved kinetic energy is not conserved.



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