A 0.300 kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.215 kg puck that is initially moving along the x axis with a velocity of 2.14 m/s. After the collision, the 0.215 kg puck has a speed of 1.14 m/s at an angle of q = 53.5 deg to the positive x axis. Determine the velocity of the 0.300 kg puck after the collision.
Calculate the magnitude of the angle with respect to the x-axis.
Find the percentage of kinetic energy lost in the collision. Do not enter units.
Please help me solve it
To find the velocity of the 0.300 kg puck after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocity of the 0.300 kg puck as v1 and the initial velocity of the 0.215 kg puck as v2. The final velocity of the 0.215 kg puck is given as 1.14 m/s at an angle of 53.5 degrees with respect to the positive x-axis.
The momentum before the collision is given by:
Initial momentum = m1 * v1 + m2 * v2
The momentum after the collision is given by:
Final momentum = m1 * v1f + m2 * v2f
Since the system is frictionless, no external forces act on the puck, so the total momentum is conserved:
Initial momentum = Final momentum
This can be written as:
(m1 * v1 + m2 * v2) = (m1 * v1f + m2 * v2f)
Now, let's substitute the known values into the equation:
(0.300 kg * v1 + 0.215 kg * 2.14 m/s) = (0.300 kg * v1f + 0.215 kg * 1.14 m/s * cos(53.5 deg))
Simplifying the equation further:
0.300 kg * v1 + 0.215 kg * 2.14 m/s = 0.300 kg * v1f + 0.215 kg * 1.14 m/s * cos(53.5 deg)
Now, solve the equation for v1f, the velocity of the 0.300 kg puck after the collision.
In order to find the magnitude of the angle with respect to the x-axis, we can use the trigonometric relationship between the x and y components of the velocity. The magnitude of the angle q can be found using the formula:
tan(q) = (v2f * sin(q))/(v2f * cos(q))
Rearranging the equation, we have:
tan(q) = sin(q)/cos(q)
Taking the inverse tangent of both sides, we get:
q = atan(sin(q)/cos(q))
Using the given values, we can substitute them into the equation to find the value of q, the angle with respect to the x-axis.
To find the percentage of kinetic energy lost in the collision, we need to compare the initial kinetic energy with the final kinetic energy. The initial kinetic energy is given by:
Initial KE = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
The final kinetic energy is given by:
Final KE = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
The percentage of kinetic energy lost can be calculated using the formula:
Percentage KE lost = ((Initial KE - Final KE) / Initial KE) * 100
Using the given values, substitute them into the equations to find the percentage of kinetic energy lost in the collision.