A 0.384 kg puck, initially at rest on a horizon-
tal, frictionless surface, is struck by a 0.189 kg
puck moving initially along the x axis with
a speed of 2.7 m/s. After the collision, the
0.189 kg puck has a speed of 1.7 m/s at an
angle of 28� to the positive x axis.
Determine the velocity of the 0.384kg puck
after the collision.
Answer in units of m/s
Linear momentum must be preserved. You do not have to assunme an elastic collision. Total momentum remains zero along the y direction. That will tell you the y momentum and velocity component of the 0.384 kg puck.
0.189*1.7 sin 28 = -0.384 Vy
Solve for Vy.
For the x component, Vx, require total momentum conservation in the x direction
0.189*2.7 = 0.189*1.7*cos28 + 0.384*Vx
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the initial momentum of the system is given by the sum of the individual momenta of the two pucks:
Initial momentum = (mass1 * velocity1) + (mass2 * velocity2)
Let's identify the values given in the problem:
mass1 = 0.384 kg (mass of the 0.384 kg puck)
mass2 = 0.189 kg (mass of the 0.189 kg puck)
velocity1 = 0 m/s (initial velocity of the 0.384 kg puck, which is at rest)
velocity2 = 2.7 m/s (initial velocity of the 0.189 kg puck along the x-axis)
Now let's calculate the initial momentum:
Initial momentum = (0.384 kg * 0 m/s) + (0.189 kg * 2.7 m/s)
= 0 + 0.5113 kg·m/s
= 0.5113 kg·m/s
After the collision, the 0.384 kg puck will have a final velocity. Let's call this velocity V. The 0.189 kg puck will have a final velocity of 1.7 m/s at an angle of 28 degrees to the positive x-axis.
Using the principle of conservation of momentum again, the final momentum of the system is given by:
Final momentum = (mass1 * final velocity1) + (mass2 * final velocity2)
Final momentum = (0.384 kg * V) + (0.189 kg * 1.7 m/s)
The final momentum should be equal to the initial momentum, so we can set up the equation:
0.5113 kg·m/s = (0.384 kg * V) + (0.189 kg * 1.7 m/s)
Now we can solve this equation to find the value of V:
0.5113 kg·m/s - (0.189 kg * 1.7 m/s) = 0.384 kg * V
0.5113 kg·m/s - 0.3213 kg·m/s = 0.384 kg * V
0.19 kg·m/s = 0.384 kg * V
V = (0.19 kg·m/s) / 0.384 kg
V = 0.4948 m/s
So, the velocity of the 0.384 kg puck after the collision is approximately 0.4948 m/s.
To determine the velocity of the 0.384 kg puck after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Let's denote the velocity of the 0.384 kg puck as V1 (unknown), and the velocity of the 0.189 kg puck as V2 (1.7 m/s at an angle of 28 degrees to the positive x-axis).
The initial momentum before the collision (along the x-axis) is given by:
P_initial = m1 * V1_initial + m2 * V2_initial
Since the 0.384 kg puck is initially at rest, V1_initial = 0. Therefore:
P_initial = m2 * V2_initial
The final momentum after the collision (along the x-axis) is given by:
P_final = m1 * V1_final + m2 * V2_final
According to the conservation of momentum, P_initial = P_final:
m2 * V2_initial = m1 * V1_final + m2 * V2_final
Plugging in the given values:
0.189 kg * (2.7 m/s) = 0.384 kg * V1_final + 0.189 kg * (1.7 m/s) * cos(28 degrees)
Simplifying the equation:
0.5113 kg·m/s = 0.384 kg * V1_final + 0.2454 kg·m/s
Rearranging the equation to solve for V1_final:
V1_final = (0.5113 kg·m/s - 0.2454 kg·m/s) / 0.384 kg
Calculating the result:
V1_final ≈ 0.5601 m/s
Therefore, the velocity of the 0.384 kg puck after the collision is approximately 0.5601 m/s.