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dedylja [7]
4 years ago
12

Two objects of the same mass move along the same line in opposite directions. The first mass is moving with speed . The objects

collide, stick together, and move with speed 0.100 in the direction of the velocity of the first mass before the collision. What was the speed of the second mass before the collision?
0.00
0.800
0.900
10.0
1.20
and show your work please
Physics
1 answer:
Alexandra [31]4 years ago
6 0

Explanation:

Let the mass of two objects be m. Both objects move along the same line in opposite directions. Let v and u₂ are speeds of both objects before collision.

After the collision, both objects stick together and move with the speed of 0.1 V the direction of the velocity of the first mass before the collision.

Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)v

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{m(u_1+u_2)}{(m_1+m_2)}

0.1v=\dfrac{m(v+u_2)}{(2m)}

On solving above equation, u_2=-0.8v

So, the speed of the second mass before the collision is 0.8 v. The negative sign shows that the it moves in opposite direction. Hence, this is the required solution.

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What is the force on an object with a mass of 25 kg and an acceleration of 5 m/s/s? 1 point
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The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.
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Answer:

(a) the particle position = 135 m

(b) the velocity of the particle = 44 m/s

(c) the acceleration of the particle = 50 m/s²

Explanation:

Solution to Question 2.

Given;

velocity of a particle, v = 2 - 4t + 2t³

initial position at t = 0, s₀ = 3 m

(a) the particle position at t = 3 s

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(b) the velocity of the particle at t = 3 s

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(c) the acceleration of the particle at t = 3s

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