A 0.408 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.0886 kg puck moving initially along the x axis with a speed of 3.14 m/s. After the collision, the 0.0886 kg puck has a speed of
2.14 m/s at an angle of 38◦ to the positive x axis.
Determine the velocity of the 0.408 kg puck after the collision.
Answer in units of m/s
To find the velocity of the 0.408 kg puck after the collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system remains constant before and after a collision, provided no external forces act on the system.
Mathematically, the principle of conservation of momentum can be expressed as:
m1*v1i + m2*v2i = m1*v1f + m2*v2f
Where:
m1 and m2 are the masses of the two pucks
v1i and v2i are the initial velocities of the pucks
v1f and v2f are the final velocities of the pucks
In this case, the 0.408 kg puck has a mass of m1 = 0.408 kg and an initial velocity of v1i = 0 m/s. The 0.0886 kg puck has a mass of m2 = 0.0886 kg and an initial velocity of v2i = 3.14 m/s.
Let's assume that the final velocity of the 0.408 kg puck is v1f and the final velocity of the 0.0886 kg puck is v2f.
Using the principle of conservation of momentum, we can rewrite the equation as:
(0.408 kg)(0 m/s) + (0.0886 kg)(3.14 m/s) = (0.408 kg)(v1f) + (0.0886 kg)(2.14 m/s * cos(38°))
Simplifying the equation, we have:
0.2776 kg*m/s = 0.408 kg*v1f + 0.05425 kg*m/s
Substituting the given values and solving for v1f, we get:
v1f = (0.2776 kg*m/s - 0.05425 kg*m/s) / 0.408 kg
v1f = 0.629 m/s
Therefore, the velocity of the 0.408 kg puck after the collision is 0.629 m/s.