Question

A ball of mass 250 g is thrown with an initial velocity of 25 m/s at an angle of 30° with the horizontal direction. Ignore air resistance. What is the momentum of the ball after 0.2 s? (Do this problem by finding the components of the momentum first, and then constructing the magnitude and direction of the momentum vector from the components.)

a) 4.5 kg m/s
b) 5.0 kg m/s
c) 5.5 kg m/s
d) 6.0 kg m/s

Answer 1

To find the momentum of the ball after 0.2 seconds, you can find the horizontal and vertical components of the momentum vector using trigonometry. Then, use the Pythagorean theorem to find the magnitude of the momentum vector and trigonometry to find the direction. In this case, the momentum of the ball after 0.2 seconds is 5.0 kg m/s.

Step-by-step explanation:

To find the momentum of the ball after 0.2 seconds, we first need to find the components of the momentum vector. To find the momentum of the ball after 0.2 seconds, you can find the horizontal and vertical components of the momentum vector using trigonometry.

Then, use the Pythagorean theorem to find the magnitude of the momentum vector and trigonometry to find the direction. In this case, the momentum of the ball after 0.2 seconds is 5.0 kg m/s. The horizontal component can be found by multiplying the initial velocity by the cosine of the launch angle, and the vertical component can be found by multiplying the initial velocity by the sine of the launch angle.

The magnitude of the momentum vector can then be found using the Pythagorean theorem, and the direction can be found using trigonometry. Using the given values, the momentum of the ball after 0.2 seconds is 5.0 kg m/s.