A bumper car with mass m1 = 109 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 83 kg is moving to the left with a velocity of v2 = -3.5 m/s. The two cars have an elastic collision. Assume the surface is frictionless.
a. What is the velocity of the center of mass of the system?
b. What is the initial velocity of car 1 in the center-of-mass reference frame?
c. What is the final velocity of car 1 in the center-of-mass reference frame?
d. What is the final velocity of car 1 in the ground (original) reference frame?
e. What is the final velocity of car 2 in the ground (original) reference frame?
f. In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.
What is the final speed of the two bumper cars after the collision?
g. Compare the loss in energy in the two collisions:
(i) |ΔKEelastic| = |ΔKEinelastic|
(ii) |ΔKEelastic| > |ΔKEinelastic|
(iii) |ΔKEelastic| < |ΔKEinelastic|
a. velocity of COM=[mᵢvᵢ + mᵢᵢvᵢᵢ] / [mᵢ + mᵢᵢ]
= [109(4.9) + 83(-3.5)] / (109+83)
= 1.27
b. initial velocity of car 1 = vᵢ - velocity of COM
= 4.9-1.27
= 3.63
c. final velocity of car 1 = -3.63
d. final velocity of car 1 GRF = final velocity of car 1 + velocity of COM
= -3.63 + (1.27)
= -2.36
e. final velocity of car 2 GRF = final velocity of car 2 + velcoity of COM
initial velocity of car 2 = -3.5 - 1.27 = -4.77
final velocity of car 2 = 4.77
= 4.77 + 1.27
= 6.04
f. 1.27
g. option (i) is correct
a. The velocity of the center of mass of the system can be found using the formula:
Vcm = (m1 * v1 + m2 * v2) / (m1 + m2)
Vcm = (109 kg * 4.9 m/s + 83 kg * (-3.5 m/s)) / (109 kg + 83 kg)
Vcm = (534.1 kg m/s - 290.5 kg m/s) / 192 kg
Vcm = 243.6 kg m/s / 192 kg
Vcm ≈ 1.27 m/s to the right
b. In the center-of-mass reference frame, the initial velocity of car 1 would be its velocity relative to the center of mass, which is simply:
v1_cm = v1 - Vcm
v1_cm = 4.9 m/s - 1.27 m/s
v1_cm ≈ 3.63 m/s to the right
c. Since the collision is elastic, the final velocity of car 1 in the center-of-mass reference frame would be the same as its initial velocity:
v1_cm_final = v1_cm = 3.63 m/s to the right
d. To find the final velocity of car 1 in the ground reference frame, we need to convert it from the center-of-mass reference frame to the ground reference frame. Since the center of mass is moving to the right, the final velocity would be:
v1_final = v1_cm_final + Vcm
v1_final = 3.63 m/s + 1.27 m/s
v1_final ≈ 4.90 m/s to the right
e. Similar to car 1, the final velocity of car 2 in the ground reference frame would be:
v2_final = v2_cm_final + Vcm
Since the center of mass is moving to the right, the final velocity would be:
v2_final = -3.5 m/s + 1.27 m/s
v2_final ≈ -2.23 m/s to the left
f. In an inelastic collision, the two cars latch together and move as a single unit. In this case, the final speed can be found using the conservation of momentum:
m1 * v1_initial + m2 * v2_initial = (m1 + m2) * v_final
(109 kg * 4.9 m/s) + (83 kg * (-3.5 m/s)) = (109 kg + 83 kg) * v_final
(534.1 kg m/s - 290.5 kg m/s) = (192 kg) * v_final
243.6 kg m/s = 192 kg * v_final
v_final = 243.6 kg m/s / 192 kg
v_final ≈ 1.27 m/s
So, the final speed of the two latched bumper cars is approximately 1.27 m/s.
g. (ii) |ΔKEelastic| > |ΔKEinelastic|
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
a. The velocity of the center of mass can be found using the formula:
Center of Mass Velocity = (m1 * v1 + m2 * v2) / (m1 + m2)
Substituting the given values, we get:
Center of Mass Velocity = (109 kg * 4.9 m/s + 83 kg * (-3.5 m/s)) / (109 kg + 83 kg)
Center of Mass Velocity = 562.67 kg.m/s / 192 kg
Center of Mass Velocity = 2.93 m/s (to the right)
b. In the center-of-mass reference frame, the initial velocity of car 1 would be equal to the negative of the center of mass velocity, as both cars are moving in opposite directions.
Initial velocity of car 1 = -2.93 m/s
c. In an elastic collision, the relative velocity between the two objects before and after the collision remains constant. Therefore, the final velocity of car 1 in the center-of-mass reference frame would be the negative of the initial velocity of car 2.
Final velocity of car 1 = -v2 = 3.5 m/s (to the left)
d. The final velocity of car 1 in the ground (original) reference frame can be found by adding the velocity of the center of mass to the final velocity of car 1 in the center-of-mass reference frame.
Final velocity of car 1 in the ground reference frame = Center of Mass Velocity + Final velocity of car 1 in the center-of-mass reference frame
Final velocity of car 1 in the ground reference frame = 2.93 m/s + 3.5 m/s = 6.43 m/s (to the left)
e. The final velocity of car 2 in the ground reference frame can be found by subtracting the velocity of the center of mass from the final velocity of car 1 in the ground reference frame.
Final velocity of car 2 in the ground reference frame = Final velocity of car 1 in the ground reference frame - Center of Mass Velocity
Final velocity of car 2 in the ground reference frame = 6.43 m/s - 2.93 m/s = 3.5 m/s (to the left)
f. In an inelastic collision, the two cars latch together after the collision, so they move together as a single object. The total mass of the system after the collision is the sum of the masses of car 1 and car 2.
Total mass = m1 + m2 = 109 kg + 83 kg = 192 kg
Using the principle of conservation of momentum, the initial momentum of the system is equal to the final momentum of the system.
Initial momentum of the system = Final momentum of the system
(m1 * v1) + (m2 * v2) = Total mass * final velocity (after collision)
(109 kg * 4.9 m/s) + (83 kg * (-3.5 m/s)) = 192 kg * final velocity
536.1 kg.m/s - 290.5 kg.m/s = 192 kg * final velocity
245.6 kg.m/s = 192 kg * final velocity
final velocity = 245.6 kg.m/s / 192 kg
final velocity = 1.28 m/s (to the right)
g. Comparing the loss in energy in the two collisions:
(i) The magnitude of the change in kinetic energy in an elastic collision is given by:
|ΔKEelastic| = |(1/2) * m1 * (v1^2) + (1/2) * m2 * (v2^2) - (1/2) * m1 * (final velocity of car 1)^2 - (1/2) * m2 * (final velocity of car 2)^2|
Substituting the given values, we can calculate the magnitude of ΔKEelastic.
(ii) The magnitude of the change in kinetic energy in an inelastic collision is given by:
|ΔKEinelastic| = |(1/2) * (m1 + m2) * (initial velocity of the center of mass)^2 - (1/2) * (m1 + m2) * (final velocity after collision)^2|
Substituting the given values, we can calculate the magnitude of ΔKEinelastic.
Then, we can compare the magnitudes of |ΔKEelastic| and |ΔKEinelastic|:
(i) If |ΔKEelastic| = |ΔKEinelastic|, energy is conserved in both collisions.
(ii) If |ΔKEelastic| > |ΔKEinelastic|, less energy is lost in the elastic collision compared to the inelastic collision.
(iii) If |ΔKEelastic| < |ΔKEinelastic|, more energy is lost in the elastic collision compared to the inelastic collision.