Two pucks moving on a friction less air table are about to collide. The 1.5 kg puck is moving directly east at 2 m/s. the 4 kg puck is moving directly north at 1 m/s. What is the total kinetic energy of the two-puck system before the collision?
a) sqrt(13) J
b)5 J
c)7 J
d)10 J
e)11 J
I got b) as the answer by using KE=1/2 mv^2 of both of the pucks. Then I added the two numbers up and got b).
(1/2)[(1.5x4) + (4x1)] = 5
You got it!
Yes, you got it right! To find the total kinetic energy of the two-puck system before the collision, you need to use the formula for kinetic energy, which is KE = 1/2 mv^2.
For the 1.5 kg puck moving directly east at 2 m/s, you calculate its kinetic energy as KE1 = 1/2 * 1.5 * (2^2) = 3 J.
For the 4 kg puck moving directly north at 1 m/s, you calculate its kinetic energy as KE2 = 1/2 * 4 * (1^2) = 2 J.
To find the total kinetic energy of the system, you add the kinetic energies of the two pucks: KE_total = KE1 + KE2 = 3 J + 2 J = 5 J.
Therefore, the correct answer is b) 5 J. Well done!