Question

A basketball (m = 0.60 kg) is dropped from rest. Just before striking the floor, the magnitude of the basketball's momentum is 3.1 kg-m/s. What is the magnitude of the ball's velocity, just before hitting the floor? | 1 a. 2.3 m/s (2b. 3.2 m/s (3) c. 5.2 m/s 4 d. 10 m/s © e. 27 m/s

Answer 1

Question 1

Given mass m = 0.6 kg and momentum p = 3.1 kg m/s

We have, Momentum = mass * velocity

p = m * v

Therefore, velocity, v = p / m

v = 3.1 / 0.6

v = 5.2 m/s

Answer is option c.

Question 2

Given mass m = 0.6 kg and momentum p = 3.1 kg m/s

Kinetic energy just before striking the floor

  $\\ K = \frac{p^2}{2m} \\ \\ K = \frac{3.1^2}{2 * 0.6} \\\\ K = 8.01 \ J$

By conservation of mechanical energy,

Potential energy at height h = Kinetic energy just before striking the floor

Therefore , height from which it is dropped is

$\\h = \frac{8.01}{m * g} \\\\h = \frac{8.01}{0.6 * 9.8} \\\\h = \frac{8.01}{5.88} \\\\h = 1.362 \ m \\\\h \approx 1.4 \ m$

Answer is option c.

Question 3

Conservation of linear momentum can be applied when there is no external force acting on the body. Therefore, in the absence of an external force, a body's linear momentum remains a constant.

Answer is option b.

Question 4 and Question 5 cannot be done as their figures and graphs are not given

Question 6

During a collision, the kinetic energy is conserved when there is no deformation of the bodies or when they do not stick together after collision, that is, kinetic energy is conserved only if it is an elastic collision

Answer is option d