Question

evaluate the known quantities in the conservation equation. (cont.)
after the collision, both pucks travel together with a speed of 1.85 m/s in a direction 28.0° above the negative x - axis, 152° from the positive x - axis. the final y - component of their common velocity is then
v_{yf}=(1.85m/s)(sin152^{circ})=(1.85m/s)(sin28.0^{circ}) = 0.869m/s.
compiling these results, we know that the masses of both pucks are equal, ( m = m = 1.90\times10^{-2}kg ). the initial x - and y - component velocities of puck 1 are ( v_{xi}=0m/s ) and ( v_{yi}=1.05m/s ), respectively. we have calculated the final x - and y - component velocities of both pucks to be, respectively,
v_{xf}=v_{xf}=(1.85m/s)(cos152^{circ})=-1.63m/s
and
v_{yf}=v_{yf}=(1.85m/s)(sin152^{circ}) = 0.869m/s.
substitute known quantities and solve for the unknown quantities.
what is the x - component of the initial velocity of puck 2?
m/s

Explanation:

Step1: Apply conservation of momentum in x - direction

The conservation of momentum equation in the x - direction is \(Mv_{x1}+mv_{x2}=(M + m)v_{xf}\). Since \(M = m\), the equation becomes \(v_{x1}+v_{x2}=2v_{xf}\).

Step2: Substitute known values

We know that \(v_{x1}=0\ m/s\) and \(v_{xf}=- 1.63\ m/s\). Substituting these values into the equation \(v_{x1}+v_{x2}=2v_{xf}\), we get \(0 + v_{x2}=2\times(-1.63)\).

Answer:

\(-3.26\ m/s\)