Just checking my answers

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?
mv + mv = (m+m)v
(0.3 kg)(4 m/s) + (0.5 kg)(-4 m/s) = (0.3 + 0.5 kg) v
v = -1 m/s
ΣFt = mΔv
F (0.005 s) = (0.5 kg)(-1 m/s - -4 m/s)
F = -500 N
Is this correct.

Yes, your calculations are correct.

To determine the final velocity of the two cars after collision, you used the principle of conservation of momentum. You added the momentum of the two cars before collision (m1v1 + m2v2) and set it equal to the total momentum after collision [(m1 + m2) * v], where m1 and m2 are the masses of the cars, and v1 and v2 are their initial velocities. By plugging in the values, you correctly found the final velocity to be -1 m/s.

To calculate the force applied by the cars on each other during the collision, you used the impulse-momentum principle. The change in momentum (Δp) is equal to the force (F) multiplied by the time interval (t). You rearranged the equation to solve for force (F) and plugged in the given values to get a force of -500 N. The negative sign indicates that the force is in the opposite direction of the initial motion.

Overall, your calculations are correct, and you arrived at the final velocity and force applied by the cars on each other accurately.