A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>
Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.
∴
∴
∴ ₁₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
Can learn more about elastic collision from brainly.com/question/12644900
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Answer:
humans and whales are both mammals
humans and whales can breathe air
Answer:
The linear speed of scratch is 45.55 cm/dec
Explanation:
The scratch located from the center at radial distance = 14.5 cm.
Each minute the skip of scratch when disco record played = 30 times
Since, the scratch on the disc start revolving when disc is played and in one minute the scratch revolve 30 times. Therefore, the revolution = 30 RPM
Below is the calculation of linear speed of the scratch.
When the object is at rest, there is a zero net force due the cancellation of the object's weight <em>w</em> with the normal force <em>n</em> of the table pushing up on the object, so that by Newton's second law,
∑ <em>F</em> = <em>n</em> - <em>w</em> = 0 → <em>n</em> = <em>w</em> = <em>mg</em> = 112.5 N ≈ 113 N
where <em>m</em> = 12.5 kg and <em>g</em> = 9.80 m/s².
The minimum force <em>F</em> needed to overcome <u>maximum</u> static friction <em>f</em> and get the object moving is
<em>F</em> > <em>f</em> = 0.50 <em>n</em> = 61.25 N ≈ 61.3 N
which means a push of <em>F</em> = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.