A 4.00 kg object is moving at 5.00 m/s NORTH. It strikes a 6.00 kg object that is moving WEST at 2.00 m/s. The objects undergo a perfectly inelastic collision (stick together). The velocity of the 4.00 kg object after the collision is
A) 2.54 m/s at an angle of 35.0 degrees NORTH of WEST.
B) 1.93 m/s at an angle of 59.0 degrees NORTH of WEST.
C) 1.93 m/s at an angle of 45.0 degrees NORTH of WEST.
D) 2.33 m/s at an angle of 59.0 degrees NORTH of WEST.
E) 2.33 m/s at an angle of 45.0 degrees NORTH of WEST.

Question

Quizplus · Accepted Answer

Before the collision, we can break down the velocities into their x and y components:
Object 1 (4.00 kg): vx1 = 0 m/s, vy1 = 5.00 m/s
Object 2 (6.00 kg): vx2 = -2.00 m/s, vy2 = 0 m/s

Conservation of momentum tells us that the momentum in the x direction is conserved:
(m1)(vx1) + (m2)(vx2) = (m1 + m2)(vxf)
where vxf is the final velocity in the x direction after the collision.

Plugging in numbers, we get:
(4.00 kg)(0 m/s) + (6.00 kg)(-2.00 m/s) = (4.00 kg + 6.00 kg)(vxf)
vxf = -1.20 m/s

Since the two objects stick together after the collision, we can find their combined mass and treat them as a single object:
m = m1 + m2 = 10.0 kg

Conservation of momentum in the y direction tells us:
(m1)(vy1) + (m2)(vy2) = m(vyf)
where vyf is the final velocity in the y direction after the collision.

Plugging in numbers:
(4.00 kg)(5.00 m/s) + (6.00 kg)(0 m/s) = (10.0 kg)(vyf)
vyf = 2.00 m/s

Now we can combine the x and y components to find the magnitude and direction of the final velocity. Using the Pythagorean theorem:
vf = sqrt(vxf^2 + vyf^2) = 2.33 m/s

And using trigonometry:
theta = tan^-1(vyf/vxf) = 59.0 degrees NORTH of WEST.

So the answer is D) 2.33 m/s at an angle of 59.0 degrees NORTH of WEST.

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