posted by Joe on .
An air-hockey puck of mass m floats across the table essentially frictionless on a cushion of
air. This puck bumps nearly head-on into a second puck that has 3 times the mass, moving in the
opposite direction but with the same speed, v i , as the lighter puck. Immediately after the
collision, the first puck is moving at a right angle to its original direction. What is the final speed
of this puck? What is the final velocity of the heavier puck? (That means speed and direction…)
Answers will contain m and v i and some angle θ.
Draw the figure (origin is at the origin of the coordinate system):
Vector p1=m1•v1=m•v is directed to the right along +x-axis,
vector p2=m2•v2=3•m•v is directed to the left in –x –direction
Vector sum of the vectors p1 and p2 is directed to the left and =|2•p1|=|2•m•v|;
Vector p'1=m1•u1=m•u1 is directed downwards along –y-axis,
vector p'2=m2•u2=3m•u2 is directed upwards to the left and makes an angle θ with negative direction of x-axis.
Vector sum of the vectors p'1 and p'2 is equal to the vector sum of vectors p1 and p2 according to the law of conservation of linear momentum. Due to this law x- and y- projections of the vectors of momentums are
x: p1- p2 = 0 - p'2•cos θ,
y: 0 = - p'1 + p'2•sin θ.
m•v-3•m•v =3•m•u2•cos θ,
0= - m•u1+3•m•u2•sin θ.
-2• v =3• u2•cos θ,
0= - u1+3• u2•sin θ.
u1 =3•u2• sin θ =3•2•v•sin θ/3•cos θ =2•v•tan θ.