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:
part a
Using conservation of momentum:
part b
Using energy conservation:
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
Considering the second experiment
initial kinetic energy
Final kinetic energy
Loss in kinetic energy
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
Considering the second experiment
initial kinetic energy
Final kinetic energy
Loss in kinetic energy
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.