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

Question

Quizplus · Accepted Answer

Since momentum is conserved in an inelastic collision, we can set up the following equation:
(momentum before collision) = (momentum after collision)
(m1)(v1) + (m2)(v2) = (m1 + m2)(vf)
where m1 and v1 are the mass and velocity of the 4.00 kg object, m2 and v2 are the mass and velocity of the 6.00 kg object before the collision, and vf is the velocity of the two objects stuck together after the collision.

Plugging in the given values:
(4.00 kg)(5.00 m/s) + (6.00 kg)(2.00 m/s) = (4.00 kg + 6.00 kg)(vf)
20.00 kg•m/s + 12.00 kg•m/s = 10.00 kg•vf
32.00 kg•m/s = 10.00 kg•vf
vf = 3.20 m/s

Now we need to find the direction of vf. Since the 4.00 kg object was initially moving north and the 6.00 kg object was initially moving east, the two directions are perpendicular. After the collision, the two objects are stuck together and moving in a diagonal direction somewhere between north and east.

We can use trigonometry to find the direction. In the right triangle formed by vf, the x-axis (east), and the y-axis (north), the angle θ between vf and the x-axis is:

tan θ = (vf in the y-direction) / (vf in the x-direction)
tan θ = (4.00 kg • 5.00 m/s) / (10.00 kg • 3.20 m/s)
tan θ = 0.625
θ = 31.2°

Since θ is the angle between vf and the x-axis (east), the angle we want is 90° - θ, or:

90° - 31.2° = 58.8°

So the velocity of the 6.00 kg object after the collision is 2.33 m/s at an angle of 58.8° north of east, which is answer choice B.

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