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 have a completely inelastic (stick together)collision.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

To solve this problem, we need to use conservation of momentum in both x and y directions.

Initial momentum in x direction = 4.00 kg * 5.00 m/s (NORTH) + 6.00 kg * 2.00 m/s (WEST)
= 20.00 kg*m/s (WEST)

Final momentum in x direction = (4.00 kg + 6.00 kg) * Vx (North-west)
= 10.00 kg * Vx (North-west)

By conservation of momentum, we have:

20.00 kg*m/s (WEST) = 10.00 kg * Vx (North-west)

Solving for Vx, we get:

Vx = 2.00 m/s (WEST)

Similarly, we can use conservation of momentum in y direction to find Vy, which is:

Vy = 3.00 m/s (NORTH)

Therefore, the velocity of the 4.00 kg object after the collision is:

V = sqrt(Vx^2 + Vy^2) = 2.33 m/s

The direction of velocity can be found by using the inverse tangent function:

theta = tan^-1(Vy/Vx) = 59.0 degrees NORTH of WEST.

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

A 400 Kg Object Is Moving at 5