Question

question 3
5 points
when two masses collide and stick together, their kinetic energy is conserved.
a true
b false
clear selection
question 4
5 points
you fire a projectile horizontally from a table top 1 m above the floor. if the projectile travels 2m horizontally before hitting the floor,
how fast was the projectile traveling when fired (in m/s)?
add your answer

Explanation:

Question 3

When two masses collide and stick together, it is a perfectly inelastic collision. In a perfectly inelastic collision, momentum is conserved, but kinetic energy is not conserved. Some of the kinetic energy is lost as heat, sound, or due to deformation of the colliding objects.

Step1: Find the time of flight

The vertical motion of the projectile is a free - fall motion. The vertical displacement \(y = 1m\), and the equation for vertical displacement in free - fall is \(y=v_{0y}t+\frac{1}{2}gt^{2}\). Since the projectile is fired horizontally, \(v_{0y} = 0\). So, \(y=\frac{1}{2}gt^{2}\), and \(t=\sqrt{\frac{2y}{g}}\).
Substituting \(y = 1m\) and \(g = 9.8m/s^{2}\), we get \(t=\sqrt{\frac{2\times1}{9.8}}\approx\sqrt{\frac{2}{9.8}}\approx0.45s\)

Step2: Find the initial horizontal velocity

The horizontal motion of the projectile is a uniform motion (because there is no horizontal acceleration, \(a_x = 0\)). The horizontal displacement \(x = 2m\), and the equation for horizontal displacement is \(x = v_{0x}t\). Then \(v_{0x}=\frac{x}{t}\)
Substituting \(x = 2m\) and \(t\approx0.45s\), we get \(v_{0x}=\frac{2}{0.45}\approx4.44m/s\)

Answer:

B. False

Question 4