A 700-kg car stopped at an intersection is rear-ended by a 1710-kg truck moving with a speed of 16.0m/s .
1) If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck.
2)Find the final speed of the car.
1) Oh boy, this sounds like a big "rear"tangle! Well, in an approximately elastic collision, we can use the conservation of momentum to solve this one. The total momentum before the collision is equal to the total momentum after the collision.
The momentum of the car before the collision is zero since it's in neutral and not moving. So the momentum of the truck before the collision is:
Momentum_before = mass_truck * velocity_truck = 1710 kg * 16.0 m/s
To find the final speed of the truck, we need to look at the momentum after the collision. Since it's an elastic collision, the total momentum after the collision is also equal to zero.
Momentum_after = mass_truck * final_velocity_truck = 0
Solving for final_velocity_truck, we have:
final_velocity_truck = 0 / 1710 kg
So, the final speed of the truck is a big ol' zero! It didn't go anywhere!
2) Now, onto the final speed of the car. We use the same idea of conservation of momentum here.
The momentum of the truck before the collision is:
Momentum_before = mass_truck * velocity_truck = 1710 kg * 16.0 m/s
The momentum of the car before the collision is:
Momentum_before = mass_car * velocity_car = 700 kg * 0 m/s (since it's stopped)
The total momentum before the collision is:
Total_momentum_before = Momentum_truck_before + Momentum_car_before
To find the final speed of the car, we consider the momentum after the collision. Since it's an elastic collision, the total momentum after the collision should still be zero.
Total_momentum_after = Momentum_truck_after + Momentum_car_after = 0
Now we can solve for the final velocity of the car (which is what we're after!):
final_velocity_car = - (mass_truck * velocity_truck) / mass_car
So, the negative sign indicates that the car will have a velocity in the opposite direction of the truck. You should calculate the final velocity using the given values.
To solve this problem, we can use the principles of conservation of momentum and kinetic energy.
1) Since the collision is approximately elastic, both momentum and kinetic energy are conserved.
Let's denote the initial velocity of the car as v1_i, the final velocity of the car as v1_f, the initial velocity of the truck as v2_i, and the final velocity of the truck as v2_f.
Using the conservation of momentum, we have:
(m1 * v1_i) + (m2 * v2_i) = (m1 * v1_f) + (m2 * v2_f)
where:
m1 = mass of the car = 700 kg
m2 = mass of the truck = 1710 kg
v1_i = initial velocity of the car = 0 m/s (since the car is stopped)
v2_i = initial velocity of the truck = 16.0 m/s
v1_f = final velocity of the car (to be determined)
v2_f = final velocity of the truck (to be determined)
Substituting the given values into the equation, we get:
(700 kg * 0 m/s) + (1710 kg * 16.0 m/s) = (700 kg * v1_f) + (1710 kg * v2_f)
0 + (27360 kg*m/s) = 700 kg * v1_f + 1710 kg * v2_f
27360 kg*m/s = 700 kg * v1_f + 1710 kg * v2_f
Next, we can use the conservation of kinetic energy to further equate the two sides:
(1/2) * m1 * v1_i^2 + (1/2) * m2 * v2_i^2 = (1/2) * m1 * v1_f^2 + (1/2) * m2 * v2_f^2
Again, substituting the given values and simplifying, we have:
(1/2) * 700 kg * 0 m/s + (1/2) * 1710 kg * (16.0 m/s)^2 = (1/2) * 700 kg * v1_f^2 + (1/2) * 1710 kg * v2_f^2
0 + (1/2) * 1710 kg * (256.0 m^2/s^2) = (1/2) * 700 kg * v1_f^2 + (1/2) * 1710 kg * v2_f^2
(1/2) * 1710 kg * (256.0 m^2/s^2) = (1/2) * 700 kg * v1_f^2 + (1/2) * 1710 kg * v2_f^2
Now we have a system of two equations with two unknowns (v1_f and v2_f). We can solve these equations simultaneously to find the values.
To find the final speed of the truck and the car after the collision, we can use the principles of conservation of momentum and kinetic energy.
1) Find the final speed of the truck:
In an elastic collision, both momentum and kinetic energy are conserved.
The equation for conservation of momentum is:
(mass of truck x initial velocity of truck) + (mass of car x initial velocity of car) = (mass of truck x final velocity of truck) + (mass of car x final velocity of car)
Given:
Mass of truck (m1) = 1710 kg
Initial velocity of truck (v1) = 16.0 m/s
Mass of car (m2) = 700 kg
Initial velocity of car (v2) = 0 (since the car is stopped)
Using the equation for conservation of momentum, we have:
(m1 x v1) + (m2 x v2) = (m1 x final velocity of truck) + (m2 x final velocity of car)
(1710 kg x 16.0 m/s) + (700 kg x 0) = (1710 kg x final velocity of truck) + (700 kg x final velocity of car)
From the given information, we know that the car was in neutral and its brakes were off. This means that the final velocity of the car after the collision would be equal to the final velocity of the truck. Let's call this final velocity "v".
(1710 kg x 16.0 m/s) + (700 kg x 0) = (1710 kg x v) + (700 kg x v)
Now we can solve for "v":
(27360 kg·m/s) = (2410 kg x v)
v = 27360 kg·m/s / 2410 kg
v ≈ 11.36 m/s
So, the final speed of the truck after the collision is approximately 11.36 m/s.
2) Find the final speed of the car:
Since the car was in neutral and its brakes were off, its final velocity after the collision would be equal to the final velocity of the truck, which we calculated to be approximately 11.36 m/s.
Therefore, the final speed of the car is also approximately 11.36 m/s.