Answer:it is a
Explanation hope this helps .
I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s². The solutions would be completely different if the same scenario were to play out in other places.
A ball is thrown upward with a speed of 40 m/s. Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.
So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.
Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip. After another 4.08 seconds, the ball has returned to the height of the hand which flung it. In total, the ball is in the air for <em>8.16 seconds</em> up and down.
(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then
(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>
==> <em>v</em> ≈ 6.2 m/s
(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:
<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>
<em />
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>

</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have

now we can say

so we will have


Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point