Question

1. coin 1 slides across a horizontal surface with speed v and collides with identical coin 2 that is initially at rest. immediately after the collision, coin 2 has speed 3v/4. which of the following correctly describes the speed v₁ of coin 1 immediately after the collision and how the kinetic energy of the two-coin system changes, if at all, during the collision? frictional forces are negligible.

	speed of coin	kinetic energy
a	v₁ = 0	remains constant
b	v₁ = 0	decreases
c	v₁ = \\(\frac{v}{4}\\)	remains constant
d	v₁ = \\(\frac{v}{4}\\)	decreases

Explanation:

Step1: Apply conservation of momentum

Let mass of each coin be $m$. Initial momentum: $mv + m(0) = mv$. Final momentum: $mv_1 + m(\frac{3v}{4})$. Set equal: $mv = mv_1 + \frac{3mv}{4}$ → $v_1 = \frac{v}{4}$.

Step2: Compare initial and final KE

Initial KE: $\frac{1}{2}mv^2$. Final KE: $\frac{1}{2}m(\frac{v}{4})^2 + \frac{1}{2}m(\frac{3v}{4})^2 = \frac{mv^2}{32} + \frac{9mv^2}{32} = \frac{10mv^2}{32} = \frac{5mv^2}{16} < \frac{1}{2}mv^2$. KE decreases.

Answer:

D. $v_1 = \frac{v}{4}$, Decreases