Question

A clay ball with mass 0.2 kg traveling south at 1 m/s collides with another 0.2 kg clay ball traveling east at 1.5 m/s. If the balls stick together after the collision, how fast do they move together after the collision?

Answer 1

By applying the conservation of momentum and using Pythagoras' theorem, the final velocity of two clay balls sticking together after collision is approximately √0.325 m/s in a southeast direction.

Step-by-step explanation:

In a collision where two objects stick together, we use the conservation of momentum to find their final velocity. Since momentum is a vector quantity, we need to consider the direction of each object’s momentum before the collision.

The first clay ball (mass 0.2 kg, velocity 1 m/s south) has a momentum of 0.2 kg × 1 m/s south. The second clay ball (mass 0.2 kg, velocity 1.5 m/s east) has a momentum of 0.2 kg × 1.5 m/s east. To find the total momentum, we combine these vectors. The final momentum of the system will be the vector sum of these momenta.

Using Pythagoras' theorem, we can find the magnitude of the combined momentum and thus the final velocity. The magnitude of the total momentum ptotal is:

ptotal = √(0.2 kg × 1 m/s)2 + (0.2 kg × 1.5 m/s)2 = √(0.04 + 0.09) kg2 m2/s2 = √0.13 kg m/s

The final velocity vf is then:

vf = ptotal / (total mass) = √0.13 kg m/s / 0.4 kg = √0.325 m/s

Therefore, the velocity of the two clay balls sticking together after the collision is approximately √0.325 m/s at an angle southeast, based on the direction of the individual momenta.