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Vsevolod [243]
3 years ago
6

Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b

e at rest after the collision?
Physics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not ->  total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

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3 years ago
A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a ve
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Answer:

8.1\:\mathrm{Ns}

Explanation:

The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:

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Since the basketball is changing direction, its total change in velocity is:

\Delta v = 20-(-15)=35\:\mathrm{m/s}.

Therefore, the basketball's change in momentum is:

\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}.

Thus, the impulse on the basketball is \fbox{$8.1\:\mathrm{Ns}$} (two significant figures).

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3 years ago
The speed of sound at sea level is normally about 340 m/s. A stationary fire alarm has a frequency of 15,000 Hz. An observer run
Musya8 [376]

Answer:

The velocity of the observer is 6.8 m/s

Explanation:

Doppler effect equation is given by the formula:

f'=\frac{v+v_o}{v-v_s} f\\\\where\ f'\ is \ the\ observed \ frequency=15300\ Hz,v_o\ is\ the\ velocity\ of\ the\ observer,\\v_s\ is\ the\ velocity\ of \ source=0,v\ is\ the\ velocity\ of\ sound=340\ m/s\ and\ f\ is\ actual\\frequency=15000\ Hz.\\\\substituting:\\\\15300=15000(\frac{340+v_o}{340} )\\\\340+v_o=346.8\\\\v_o=6.8\ m/sThe velocity of the observer is 6.8 m/s

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3 years ago
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