Answer:
B. changing by a constant amount each second
Explanation:
thats my answer
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.
Answer:
The ball would hit the floor approximately 
after leaving the table.
The ball would travel approximately 
horizontally after leaving the table.
(Assumption: 
.)
Explanation:
Let 
denote the change to the height of the ball. Let 
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let 
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:
.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was 
.
The height of the table was 
. Therefore, after hitting the floor, the ball would be 
below where it was before leaving the table. Hence, 
.
The equation becomes:
.
Solve for :
.
In other words, it would take approximately for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at 
) until the ball hits the floor.
The ball was in the air for approximately 
and would have travelled approximately 
horizontally during the flight.
