Object 1 (mass=6.4 kg) traveling in the +x direction at a speed of 14.01m/s undergoes an inelastic collision with Object 2 (mass=5.10kg) which is at rest. Object two flies off at a speed of 11.94m/s and an angle θ2=49.8 °. The diagram below is just a sketch and is not meant to have accurate angles. It is possible that Object 1 travels back and to the left, though it is not possible for it to travel in the -y direction. (Do you know why?)
Collision of 2-objects. Object 2 moving into fourth quadrant
a.) What is the total momentum in x direction after the collision?
b.) What is the momentum of Object 1 in the y direction after the collision?
c.)What is the speed of Object 1 after the collision?
d.) What is the angle θ1 in degrees? (Do not enter units)
You will have to do the law of conservation of momentum in two orthogonal directions, I recommend x and y.
You have two equations then, and you have two unknowns; speed of object 1 in the x, and y directions.
Get busy. A bit of algebra involved.
To answer these questions, we need to use the principles of conservation of momentum and conservation of kinetic energy.
a) To find the total momentum in the x direction after the collision, we first need to calculate the initial momentum in the x direction and the momentum of Object 2 in the x direction after the collision.
The initial momentum of Object 1 in the x direction can be calculated using the formula:
P_initial = m1 * v1_initial
where m1 is the mass of Object 1 and v1_initial is the initial velocity of Object 1 in the x direction.
Given:
m1 = 6.4 kg
v1_initial = 14.01 m/s
P_initial = 6.4 kg * 14.01 m/s = 89.664 kg*m/s
During an inelastic collision, the total momentum is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision. Since Object 2 is initially at rest, the initial momentum of Object 2 in the x direction is zero.
Therefore, the total momentum in the x direction after the collision is:
P_final = P_initial + P2_x
Since the only momentum in the x direction after the collision is the momentum of Object 2, we can write:
P_final = P2_x
b) To find the momentum of Object 1 in the y direction after the collision, we need to use the principle of conservation of kinetic energy.
In an inelastic collision, some kinetic energy is usually lost, resulting in objects sticking together or deforming. In this case, we can assume no energy is lost, so the kinetic energy is conserved.
Therefore, the momentum in the y direction after the collision would be zero, as Object 1 cannot travel in the -y direction.
c) To find the speed of Object 1 after the collision, we can use the principle of conservation of kinetic energy.
Since the kinetic energy is conserved:
KE_initial = KE_final
The initial kinetic energy is given by:
KE_initial = 1/2 * m1 * v1_initial^2
The final kinetic energy of Object 1 after the collision is given by:
KE_final = 1/2 * m1 * v1_final^2
Since the mass of Object 1 remains the same, we can equate the initial and final kinetic energies:
1/2 * m1 * v1_initial^2 = 1/2 * m1 * v1_final^2
Simplifying the equation:
v1_final^2 = (v1_initial^2 * m1) / m1
v1_final^2 = v1_initial^2
Taking the square root of both sides:
v1_final = v1_initial
Therefore, the speed of Object 1 after the collision is still 14.01 m/s.
d) To find the angle theta1 (θ1) in degrees, we can use trigonometry.
From the given information, we know that Object 2 flies off at an angle of θ2 = 49.8°. Since the collision is inelastic, Objects 1 and 2 stick together after the collision.
The angle θ1 can be calculated using the formula:
θ1 = atan(v1_final_y / v1_final_x)
Since the momentum in the y direction after the collision is zero, the y-component of the velocity of Object 1 after the collision is zero as well.
θ1 = atan(0 / v1_final_x)
θ1 = atan(0) = 0°
Therefore, the angle θ1 is 0°.