A 821-kg car stopped at an intersection is rear-ended by a 1810-kg truck moving with a speed of 11.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
To find the final speed of both vehicles after the collision, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision. Mathematically, this can be expressed as:
(m1 * u1) + (m2 * u2) = (m1 * v1) + (m2 * v2)
Where:
m1 = mass of car = 821 kg
u1 = initial velocity of car = 0 m/s (as it is stopped)
m2 = mass of truck = 1810 kg
u2 = initial velocity of truck = 11.5 m/s (given)
v1 = final velocity of car (to be determined)
v2 = final velocity of truck (to be determined)
Substituting the given values into the equation:
(821 kg * 0 m/s) + (1810 kg * 11.5 m/s) = (821 kg * v1) + (1810 kg * v2)
0 + 20815 kg*m/s = 821 kg * v1 + 1810 kg * v2
Now, we need one more equation involving the coefficients of restitution, which is given by:
e = (v2 - v1) / (u1 - u2)
Since the collision is approximately elastic, the coefficient of restitution (e) would be equal to 1.
1 = (v2 - v1) / (0 - 11.5 m/s)
Simplifying, we get:
v2 - v1 = -11.5 m/s
Rearranging this equation, we get:
v2 = v1 - 11.5 m/s
Now, we can substitute this value into the earlier equation:
0 + 20815 kg*m/s = 821 kg * v1 + 1810 kg * (v1 - 11.5 m/s)
0 + 20815 kg*m/s = 821 kg * v1 + 1810 kg * v1 - 1810 kg * 11.5 m/s
Rearranging this equation, we get:
26392.5 kg*m/s = 2595 kg * v1
Dividing both sides by 2595 kg, we get:
v1 = 26392.5 kg*m/s / 2595 kg
v1 = 10.17 m/s
Now, we can substitute this value back into our equation for v2:
v2 = v1 - 11.5 m/s
v2 = 10.17 m/s - 11.5 m/s
v2 = -1.33 m/s
Therefore, the final speed of the car after the collision is 10.17 m/s and the final speed of the truck after the collision is -1.33 m/s (indicating that it is moving in the opposite direction).
To find the final speed of both vehicles after the collision, we can use the principles of conservation of momentum and kinetic energy.
1. Conservation of momentum:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The equation for momentum is:
momentum = mass × velocity
Before the collision:
The car is stationary, so its initial momentum is zero: momentum_car_initial = 0
The truck's momentum is given by: momentum_truck_initial = mass_truck × velocity_truck_initial
After the collision:
Let final velocities of the car and the truck be v_car_final and v_truck_final, respectively.
The car's momentum after the collision is: momentum_car_final = mass_car × v_car_final
The truck's momentum after the collision is: momentum_truck_final = mass_truck × v_truck_final
According to the conservation of momentum:
momentum_car_initial + momentum_truck_initial = momentum_car_final + momentum_truck_final
0 + (mass_truck × velocity_truck_initial) = (mass_car × v_car_final) + (mass_truck × v_truck_final)
2. Conservation of kinetic energy:
In an approximately elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The equation for kinetic energy is:
kinetic energy = 0.5 × mass × velocity^2
Before the collision:
The car's kinetic energy is zero because it's stationary: KE_car_initial = 0
The truck's kinetic energy is given by: KE_truck_initial = 0.5 × mass_truck × velocity_truck_initial^2
After the collision:
The car's kinetic energy after the collision is: KE_car_final = 0.5 × mass_car × v_car_final^2
The truck's kinetic energy after the collision is: KE_truck_final = 0.5 × mass_truck × v_truck_final^2
According to the conservation of kinetic energy:
KE_car_initial + KE_truck_initial = KE_car_final + KE_truck_final
0 + (0.5 × mass_truck × velocity_truck_initial^2) = (0.5 × mass_car × v_car_final^2) + (0.5 × mass_truck × v_truck_final^2)
Now, we have two equations from conservation of momentum and conservation of kinetic energy:
1. (mass_truck × velocity_truck_initial) = (mass_car × v_car_final) + (mass_truck × v_truck_final)
2. (0.5 × mass_truck × velocity_truck_initial^2) = (0.5 × mass_car × v_car_final^2) + (0.5 × mass_truck × v_truck_final^2)
We can now solve these equations simultaneously to find the final speeds of both vehicles.
Use the eqn:
v1'=v1*(m1-m2)/(m1+m2)
M2=neutral car, its v2=0 since its at rest.
m1=truck, v1=11.5.
Then, once you have v1' you can sub it into m1v1+m2v2=m1v1'+m2v2' and find v2'.
The first eqn is a special eqn only to be used when v2=0 and the collision is completely elastic. NEVER under anyother circumstances. They are specialy derived for solely this case.
Vf2=(m1+m2)(Vo1)/m1
m1=truck; m2=car; Vo1=initial speed of truck.
mo1Vo1=mf1Vf1+mf2Vf2; substitute Vf2 to find Vf1