Answer to Question #250100 in Quantum Mechanics for kyly

Question

1. Two pucks on a horizontal air table of the same mass (m1 = m2 = 6.0 g) have attracting

magnets. They are initially positioned very far from each other so that no attraction occurs.

The two were given a push and collide. The pucks end up moving at 1.40 m/s, 32° above the

–x–axis after the collision as shown. Initially, the first puck slides with a velocity 1.11 m/s due

west.

(a) What is the initial velocity (magnitude and direction) of the second puck?

(b) What is the change in kinetic energy of the system of two pucks as a result of the

collision?

Answer 1

(a) The initial velocity can be found by momentum conservation. Remember that here momentum is conserved along x- and y- axes:

$$mv_{1x}+mv_{2x}=2mu_{x},\\ mv_{1y}+mv_{2y}=2mu_{y}.\\\space\\ -1.11+v_{2x}=2(-1.4\cos32°)→v_{2x}=1.26\text{ m/s},\\ 0+v_{2y}=2(1.4\sin32°)=1.48\text{ m/s},\\\space\\ |v_2|=\sqrt{1.26^2+1.48^2}=1.94\text{ m/s},\\\space\\ \theta=\arctan\frac{v_{2y}}{v_{2x}}=49.6°\text{ N of E}.$$

(b) The change in kinetic energy is

$$|\Delta K|=|K_f-K_i|=\bigg|\frac12m(u^2-v_1^2-v_2^2)\bigg|=0.0091\text{ J}.$$