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Fynjy0 [20]
3 years ago
14

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?
A) Yes, since momentum is conserved.
B) No, since momentum is not conserved.
C) Yes, since kinetic energy is conserved.
D) No, since kinetic energy is not conserved.
E) Not enough information.
Physics
1 answer:
TEA [102]3 years ago
5 0

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

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Answer:

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Explanation:

Let us call v_p the speed of the plane and v_w the speed of the wind. When the plane is flying against the wind, it covers the distance of 900-km in 2 hours (120 minutes); therefore;

(1). v_p - v_w = \dfrac{900km}{120min}

And when the plane is flying with the wind, it covers the same distance in 1 hour 48 minutes (108 minutes)

(2). v_p+v_w= \dfrac{900km}{108min}

From equation (1) we solve for v_p and get:

v_p = \dfrac{900km}{120min}+v_w,

and by putting this into equation (2) we get:

\dfrac{900km}{120min}+v_w+v_w= \dfrac{900km}{108min}

2v_w= \dfrac{900km}{108min}-\dfrac{900km}{120min}

2v_w = 8.3km/min - 7.5km/min

2v_w = 0.83km/min

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\boxed{v_w= 25km/hr }

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3 years ago
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Explanation:

From the ΔABC

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Answer:

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