Question #250100

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?


1
Expert's answer
2021-10-12T12:32:32-0400

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


mv1x+mv2x=2mux,mv1y+mv2y=2muy. 1.11+v2x=2(1.4cos32°)v2x=1.26 m/s,0+v2y=2(1.4sin32°)=1.48 m/s, v2=1.262+1.482=1.94 m/s, θ=arctanv2yv2x=49.6° N of E.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


ΔK=KfKi=12m(u2v12v22)=0.0091 J.|\Delta K|=|K_f-K_i|=\bigg|\frac12m(u^2-v_1^2-v_2^2)\bigg|=0.0091\text{ J}.


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