Question
Initially, mass one (2.00 kg) has a velocity of 5.80 m/s and mass two (3.00 kg) is at rest. After they collide, mass one has a velocity of 2.15 m/s at an angle theta = 33.0 degrees. What is the speed of mass two after the collision?
What is the angle phi, between mass two's velocity and the initial velocity of mass one? (Give your answer as a positive number in degrees.)
What is the ratio of final kinetic energy to initial kinetic energy in this collision? (Note, it is possible that the masses gained or lost kinetic energy in this collision. It may seem unrealistic to gain kinetic energy in a collision, but it can happen if some energy source is released during the collision.)
Explanation / Answer
Data: Mass of the first body, m1 = 2 kg Mass of the second body, m2 = 3 kg Initial velocity of the first body, v1i = 5.8 m/s Initial velocity of the second body, v2i = 0 m/s Final velocity of the first, v1f = 2.15 m/s final velocity of the second, v2f = ? Angle, = 33 deg Angle, = ? Solution: (a) According to the law of conservation of momentum: Total final momentum after collision = Total initial momentum before collision Along x-axis: m1 v1f cos + m2 v2f cos = m1 v1i + m2 v2i 2 * 2.15 * cos 33 + 3 * v2f * cos = 2 * 5.8 + 3 * 0 3.61 + 3 * v2f cos = 11.6 3 v2f cos = 7.99 v2f cos = 2.66 ...(1) Along y-axis: m1 v1f sin - m2 v2f sin = 0 m2 v2f sin = m1 v1f sin 3 v2f sin = 2 * 2.15 * sin 33 v2f sin = 0.781 ...(2) By dividing (2) with (1): tan = 0.2935 = arc tan (0.2935) = 16.36 deg So, v2f = 2.66 / cos 16.36 [ from (1) ] = 2.77 m/s Ans: Speed of the mass 2 after collision, v2f = 2.77 m/s Angle, = 16.36 deg (b) Initial kinetic energy, Ki = (1/2) m1 v1i^2 + (1/2) m2 v2i^2 = 0.5 * 2 * 5.8^2 + 0 = 33.64 J Final kinetic energy, Kf = (1/2) m1 v1f^2 + (1/2) m2 v2f^2 = 0.5 * [ 2 * 2.15^2 + 3 * 2.77^2 ] = 32.26 J Ratio of final KE and initial KE = Kf / Ki = 32.26 / 33.64 = 0.9591 Ans: Ratio of kinetic energies = 0.9591