(a) What is the speed of the four cars after the collision? m/s (b) How much mec
ID: 1413976 • Letter: #
Question
(a) What is the speed of the four cars after the collision?
m/s
(b) How much mechanical energy is lost in the collision?
J
5)A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle of = 49.0° to the positive x axis (see the figure below).
(a) Determine the velocity of the 0.300-kg puck after the collision.
m/s
(b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision.
Explanation / Answer
a)
here
momentum of 1st car = m(V1) = (2.47 * 10^4) * (4.02) = 9.9 * 10^4 kg-m/s
momentum of combined 3 cars = M(V2) = (3) * (2.47 * 10^4) * (2.01) = 14.89 * 10^4 kg-m/s
total momentum before collision = (9.9 + 14.89 ) * 10^4 = 24.8 * 10^4 kg-m/s
total mass after collision = 4(2.47 * 10^4) = 9.88 * 10^4 kg
by law of conservation of momentum:
total momentum before collision = total momentum after collision
(9.88 * 10^4) * V = 24.8 * 10^4
V = (24.8 / 9.88) = 2.51 m/s
b)
KE = 0.5 * m * v^2
KE of 1st car = (0.5) * (2.47 * 10^4) * (4.02)^2 = 19.95 *10^4 J
KE of combined 3 cars = (0.5) * (3) * (2.47 * 10^4) * (2.01)^2 = 14.96 * 10^4 J
Total KE of 4 cars BEFORE collision = (19.95 + 14.96) * 10^4 J = 34.91 * 10^4 J
KE of combined 4 cars AFTER collision = (0.5) * (9.88 * 10^4) * (2.51)^2 = 31.12 * 10^4 J
Energy lost = (34.91 - 31.12) * 10^4 J
energy lost = 3.79 * 10^4 J
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