A 56.0-kg skater is traveling due east at a speed of 2.80 m/s. A 77.0-kg skater is moving due south at a speed of 4.40 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf. Find (a) the angle and (b) the speed vf, assuming that friction can be ignored.
The momentum is conserved. Therefore the final momentum has E and S as before.
Momentum E=56*2.8
momentum S=4.4*77
Use these to get the angle
arctanTheta=MomentumS/momentumE
where theta is the angle S of E.
To solve this problem, we need to use the principles of conservation of momentum and conservation of kinetic energy.
Let's begin by using the conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the 56.0-kg skater is moving due east, so its momentum is given by 56.0 kg * 2.80 m/s = 156.8 kg·m/s toward the east direction.
Similarly, the 77.0-kg skater is moving due south, so its momentum is given by 77.0 kg * -4.40 m/s = -338.8 kg·m/s toward the south direction. Note that we assigned a negative sign to the velocity of the 77.0-kg skater because it is moving in the opposite direction of our chosen positive direction.
Let's define the x-axis as the east direction and the y-axis as the north direction. The momentum of the two skaters after they hold on to each other can be written as a vector sum:
Final momentum = (156.8 kg·m/s) * cos(theta) in the x-direction + (-338.8 kg·m/s) * sin(theta) in the y-direction
Since momentum is conserved and the two skaters hold on to each other, the final momentum will be zero in the x-direction and y-direction:
0 = (156.8 kg·m/s) * cos(theta) - (338.8 kg·m/s) * sin(theta)
Now, let's move on to conservation of kinetic energy. The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
The kinetic energy of an object is calculated by multiplying half of its mass by the square of its velocity. In this case, the total kinetic energy before the collision is:
KE_before = (1/2) * (56.0 kg) * (2.80 m/s)^2 + (1/2) * (77.0 kg) * (4.40 m/s)^2
After the collision, the two skaters hold on to each other and move off at an angle south of east with a speed of vf. The mass of the combined skaters is 56.0 kg + 77.0 kg = 133.0 kg.
The total kinetic energy after the collision is:
KE_after = (1/2) * (133.0 kg) * (vf)^2
Since kinetic energy is conserved, we can equate the two expressions for kinetic energy:
(1/2) * (56.0 kg) * (2.80 m/s)^2 + (1/2) * (77.0 kg) * (4.40 m/s)^2 = (1/2) * (133.0 kg) * (vf)^2
Now, we have two equations. We can solve them simultaneously to find a and vf.
First, solve the momentum equation for sin(theta):
(156.8 kg·m/s) * cos(theta) = (338.8 kg·m/s) * sin(theta)
sin(theta) = (156.8 kg·m/s) * cos(theta) / (338.8 kg·m/s)
Now, substitute this expression for sin(theta) into the kinetic energy equation:
(1/2) * (56.0 kg) * (2.80 m/s)^2 + (1/2) * (77.0 kg) * (4.40 m/s)^2 = (1/2) * (133.0 kg) * (vf)^2
Simplify and solve for vf:
(156.8 kg·m/s) * cos(theta) / (338.8 kg·m/s) = (vf)^2
vf = sqrt[(156.8 kg·m/s) * cos(theta) / (338.8 kg·m/s)]
Now, you can substitute any value for theta and calculate the corresponding value for vf using the last equation.
Keep in mind that the angle theta, as given in the problem statement, represents the angle south of east, which means you may need to adjust it accordingly before substituting into the equations.