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bekas [8.4K]
3 years ago
14

Puck 1 (1 kg) travels with velocity 20 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After th

e collision, puck 1 moves with a velocity of 5 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved. What is the final velocity (in m/s) of puck 2 after the collision
Physics
1 answer:
Burka [1]3 years ago
7 0

Answer:

Explanation:

Parameters given:

Mass of Puck 1, m = 1 kg

Mass of Puck 2, M = 1 kg

Initial velocity of Puck 1, u = 20 m/s

Initial velocity of Puck 2, U = 0 m/s

Final velocity of Puck 1, v = 5 m/s

Since we are told that momentum is conserved, we apply the principle of conservation of momentum:

Total initial momentum of the system = Total final momentum of the system

mu + MU = mv + MV

(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)

20 = 5 + V

V = 20 - 5 = 15 m/s

Puck 2 moves with a velocity of 15 m/s

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Does heating a cup of<br> Water allow it to dissolve more sugar ? Constant
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8 0
3 years ago
Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
soldi70 [24.7K]
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
3 years ago
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