A 0.30 kg toy car with an initial velocity of 4 m/s collides with a 0.5 kg car with an initial velocity of 4 m/s in the opposite direction. If the cars stick together during the collision, what is the final velocity of the two cars? Assume negligible friction. If the collision took 0.005 seconds, what was the force applied by the cars on one another?

Question

A 0.30 kg toy car with an initial velocity of 4 m/s collides with a 0.5 kg car with an initial velocity of 4 m/s in the opposite direction. If the cars stick together during the collision, what is the final velocity of the two cars? Assume negligible friction. If the collision took 0.005 seconds, what was the force applied by the cars on one another?

Answer 1

To find the final velocity of the two cars after collision, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum (p) of an object is given by the product of its mass (m) and velocity (v), so we can write:

p1(initial) + p2(initial) = p1(final) + p2(final)

Initially, the first car has a mass of 0.30 kg and a velocity of 4 m/s, so p1(initial) = (0.30 kg)(4 m/s).

The second car has a mass of 0.5 kg and a velocity of -4 m/s (opposite direction), so p2(initial) = (0.5 kg)(-4 m/s).

Since the two cars stick together after the collision, they have a combined mass of 0.30 kg + 0.5 kg = 0.80 kg.

Let's assume the final velocity of the combined cars is v(final).

Using the principle of conservation of momentum, we can write:

(0.30 kg)(4 m/s) + (0.5 kg)(-4 m/s) = (0.80 kg)(v(final))

Simplifying the equation:

1.2 kg m/s - 2 kg m/s = 0.80 kg v(final)

-0.8 kg m/s = 0.80 kg v(final)

Dividing both sides by 0.80 kg, we get:

v(final) = -0.8 kg m/s / 0.80 kg

v(final) = -1 m/s

Therefore, the final velocity of the two cars after the collision is -1 m/s. The negative sign indicates that the cars are moving in the opposite direction after the collision.

To calculate the force applied by the cars on each other during the collision, we can use Newton's second law of motion. It states that F (force) equals the rate of change of momentum, which can be expressed as:

F = Δp / Δt

where Δp is the change in momentum and Δt is the time interval.

The change in momentum Δp equals the final momentum minus the initial momentum, so we can write:

Δp = p(final) - p(initial)

For the first car, the change in momentum is:

Δp1 = (0.80 kg)(-1 m/s) - (0.30 kg)(4 m/s)

Δp1 = -0.80 kg m/s - 1.20 kg m/s

Δp1 = -2 kg m/s

For the second car, the change in momentum is:

Δp2 = (0.80 kg)(-1 m/s) - (0.50 kg)(-4 m/s)

Δp2 = -0.80 kg m/s - (-2 kg m/s)

Δp2 = -0.80 kg m/s + 2 kg m/s

Δp2 = 1.20 kg m/s

Since the collision took 0.005 seconds, we can substitute Δp and Δt values into the formula for force:

F = Δp / Δt

F1 = Δp1 / Δt = (-2 kg m/s) / (0.005 s) = -400 N

F2 = Δp2 / Δt = (1.20 kg m/s) / (0.005 s) = 240 N

Therefore, the force applied by the cars on each other during the collision is -400 N (for the first car) and 240 N (for the second car). The negative sign indicates that the forces are in opposite directions.