A hockey puck with mass 0.237 kg traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at 1.4 m/s strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is 30° away from the blue line at a speed of 0.79 m/s (see the figure).
What is the direction and magnitude of the velocity of the second puck after the collision?
Assume that total momentum is conserved, and solve for the speed and direction of the second puck.
To determine the direction and magnitude of the velocity of the second puck after the collision, we can use the laws of conservation of momentum and conservation of kinetic energy.
1. Conservation of Momentum:
Momentum is defined as the product of an object's mass and its velocity. According to the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Let's denote the initial velocity of the first puck along the blue line as v1i, the final velocity of the first puck as v1f, the initial velocity of the second puck as v2i, and the final velocity of the second puck as v2f.
Since the second puck is initially stationary, its initial velocity v2i is zero.
Using the conservation of momentum, we can write the equation:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f,
where m1 and m2 are the masses of the pucks (both 0.237 kg).
Substituting the given values:
0.237 kg * 1.4 m/s + 0 kg * 0 = 0.237 kg * 0.79 m/s + 0.237 kg * v2f.
Simplifying the equation:
0.3318 kg·m/s = 0.1872 kg·m/s + 0.237 kg · v2f,
0.1446 kg·m/s = 0.237 kg · v2f.
2. Conservation of Kinetic Energy:
Kinetic energy is the energy of motion. According to the law of conservation of kinetic energy, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.
The equation for kinetic energy is:
KE = (1/2) * m * v^2,
where KE is the kinetic energy, m is the mass, and v is the velocity.
Using the conservation of kinetic energy, we can write the equation:
(1/2) * m1 * v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2.
Substituting the given values:
(1/2) * 0.237 kg * (1.4 m/s)^2 + (1/2) * 0 kg * 0^2 = (1/2) * 0.237 kg * (0.79 m/s)^2 + (1/2) * 0.237 kg * v2f^2,
0.23394 kg·m^2/s^2 = 0.070168 kg·m^2/s^2 + 0.056219 kg·m^2/s^2 + 0.056219 kg·m^2/s^2,
0.23394 kg·m^2/s^2 = 0.182606 kg·m^2/s^2 + 0.056219 kg·m^2/s^2,
0.23394 kg·m^2/s^2 = 0.238825 kg·m^2/s^2.
Since the equation does not balance, it suggests that there is a loss of kinetic energy in the collision. This could be due to friction or other dissipative forces acting during the collision. Therefore, we cannot determine the value of v2f solely from the given information.
θ = 30°
x-direction:
vi + vf cos θ = x
1.4m/s - 0.79m/s(√3/2) = 0.716m/s
y-direction:
vf sin θ = y
0.79m/s * 0.5 = 0.395m/s
Speed =
√(x² + y²)
√[(0.716)² + (0.395²)] = 0.818 m/s
Direction:
Tan-¹ (y/x) = Tan-¹ (.395/.716) = 28.88° -> 28.9°