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.
u2=2•v/3cosθ,
u1 =3•u2• sin θ =3•2•v•sin θ/3•cos θ =2•v•tan θ.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's break down the problem step by step:
Step 1: Analyzing the initial situation
- The initial mass of the first puck is m.
- The initial mass of the second puck is 3m, which is three times the mass of the first puck.
- The initial speed of both pucks is vi, and they are moving in opposite directions.
- The initial velocity of the first puck is along its original direction.
Step 2: Analyzing the collision
- Since the collision is nearly head-on and the pucks experience a right-angle deflection, this indicates that the collision is elastic.
- In an elastic collision, both momentum and kinetic energy are conserved.
Step 3: Applying conservation of momentum
- Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
- Before the collision, the total momentum is given by:
p_initial = m * vi - (3m) * vi (since the second puck is moving in the opposite direction)
- After the collision, the first puck is moving at a right angle to its original direction. Therefore, only the second puck has momentum in the final direction. The momentum of the second puck is given by:
p_final = (3m) * vf * cos(θ) (where θ is the angle between the final velocity of the second puck and its initial direction)
- Since momentum is conserved, we can equate these two momenta:
m * vi - (3m) * vi = (3m) * vf * cos(θ)
Step 4: Solving for the final velocity of the first puck (vf1)
- After the collision, the final velocity of the first puck is perpendicular to its original direction.
- Therefore, vf1 = 0.
Step 5: Solving for the final velocity of the second puck (vf2)
- Rearrange the equation from Step 3 to solve for vf2:
vf2 * cos(θ) = (vi - 3vi) / 3
vf2 = (vi - 3vi) / (3 * cos(θ))
vf2 = -2vi / (3 * cos(θ))
Step 6: Determining the final speed of the first puck (vf1)
- Since the final velocity of the first puck is perpendicular to its original direction, its final speed is given by the magnitude of its final velocity.
- Therefore, vf1 = |vf1| = 0.
Step 7: Simplifying the final velocity of the second puck (vf2)
- Since the puck is moving in the opposite direction, the final velocity magnitude is the absolute value of vf2.
- Therefore, vf2 = |vf2| = 2vi / (3 * cos(θ))
So, the final speed of the first puck (vf1) is 0, and the final velocity of the second puck (vf2) is 2vi / (3 * cos(θ)) in the direction opposite to its initial direction.
Note: The angle θ is not provided in the question, so it cannot be determined without additional information. Knowing θ would be necessary to calculate the exact values.