A 645-kg object moving east at 6.25 m/s collides with a 485-kg object moving south at 4.18 m/s. The collision causes the two objects to stick together. With what velocity will the two objects be moving after the collision?Multiple choice question.3.99 m/s, 26.7° south of east5.22 m/s, 33.8° south of east5.22 m/s, 24.0° east of south3.99 m/s, 26.7° east of south

To solve this problem, we need to use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Step 1: Calculate the initial momentum of each object.

For the 645-kg object moving east: Momentum = mass * velocity = 645 kg * 6.25 m/s = 4031.25 kg*m/s (east)

For the 485-kg object moving south: Momentum = mass * velocity = 485 kg * 4.18 m/s = 2027.3 kg*m/s (south)

Step 2: The total momentum is the vector sum of the two momenta. We can treat this as a right triangle where the eastward momentum is one side, the southward momentum is another side, and the resultant momentum is the hypotenuse.

The magnitude of the total momentum (P_total) can be found using the Pythagorean theorem: P_total = sqrt((4031.25 kgm/s)^2 + (2027.3 kgm/s)^2) = 4480.8 kg*m/s

Step 3: The direction of the total momentum (θ) can be found using the tangent function: tan(θ) = opposite/adjacent = 2027.3 kgm/s / 4031.25 kgm/s θ = arctan(2027.3 kgm/s / 4031.25 kgm/s) = 26.7° south of east

Step 4: The velocity of the combined object after the collision is the total momentum divided by the total mass: Velocity = P_total / (mass1 + mass2) = 4480.8 kg*m/s / (645 kg + 485 kg) = 3.99 m/s

So, the two objects will be moving with a velocity of 3.99 m/s, 26.7° south of east after the collision. Therefore, the correct answer is "3.99 m/s, 26.7° south of east".