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QveST [7]
3 years ago
7

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before the collision, puck 1 is going 13 m/s to the east and

puck 2 is going 18 m/s to the west. After the collision, puck 1 is going 18 m/s to the west. What is the velocity of puck 2?
Physics
2 answers:
ohaa [14]3 years ago
4 0

Answer:

13 m/s east

Explanation:

We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:

p_i = p_f \\m u_1 + m u_2 = m v_1 + m v_2

where

m = 0.1 kg is the mass of each puck

u1 = +13 m/s is the initial velocity of puck 1

u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)

v1 = -18 m/s is the final velocity of puck 1

v2 = ? is the final velocity of puck 2

Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:

v_2 = u_1 + u_2 - v_1 = +13 m/s + (-18 m/s) - (-18 m/s) = +13 m/s

And the positive sign means that puck 2 is moving east.

Amanda [17]3 years ago
4 0
According to the law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
0.1 x 13 + 0.1 x 18 = 0.1 x 18 + 0.1 x v2
v2 = 13 m/s
the velocity of puck 2 is 13 m/s
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A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc
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explanation:

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3 0
2 years ago
A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
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Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

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I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

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