Answer:
<u>Inelastic collision:</u>
A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.
<u>Characteristics of an inelastic collision:</u>
- <em>the momentum of the system is conserved</em>
- <em>the momentum of the system is conservedloss of kinetic energy</em><u> </u>
<em>I</em><em>n</em><em> </em><em>a perfectly elastic collision</em><em>, the two bodies </em><em>that</em><em> </em><em>collide with each other stick together.</em>
<u>Elastic </u><u>collision</u><u>:</u>
A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.
<u>Characteristic</u><u>s</u><u> </u><u>of</u><u> </u><u>elastic</u><u> </u><u>collision</u><u>:</u>
- <em>the</em><em> </em><em>momentum</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>system</em><em> </em><em>is</em><em> </em><em>conserved</em>
- <em>no</em><em> </em><em>loss</em><em> </em><em>o</em><em>f</em><em> </em><em>kinetic</em><em> </em><em>energy</em>
In everyday life, no collision is perfectly elastic.
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ANSWER:
<u>Given examples:</u>
- Two cars colliding with each other form an example of inelastic collision.
<u>Reason:</u>
<em>(</em><em>T</em><em>hey</em><em> </em><em>lose</em><em> </em><em>kinetic</em><em> </em><em>energy</em><em> </em><em>and</em><em> </em><em>come</em><em> </em><em>to</em><em> </em><em>a</em><em> </em><em>stop</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>collision</em><em>.</em><em>)</em>
- A ball bouncing after colliding with a surface is an example of elastic collision
<u>Reason:</u>
<em>(a very less amount of kinetic energy is lost)</em>
Answer:
3secs
Explanation:
Given the following parameters
height H= 81.3m
Velocity v = 12.4m/s
Required
Time it take to reach the ground
Using the equation of motion
H = ut+1/2gt²
81.3 = 12.4t + 1/2(9.8)t²
81.3 = 12.4t + 4.9t²
4.9t² + 12.4t - 81.3 = 0
Using the general formula to find t
t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)
t = -12.4±√153.76+1593.48/2(4.9)
t = -12.4±√1747.24/9.8
t = -12.4+41.8/9.8
t = 29.4/9.8
t = 3secs
Hence it took 3secs to reach the ground
Answer:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ
