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trasher [3.6K]
3 years ago
9

13.Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it become

s zero. Thereafter, the ball begins to fall downward and attains the speed u again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?​
Physics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

Law of conservation of momentum is applicable to isolated system (no external force is applied).

In this case, the change in velocity is due to the gravitational force of earth.When the ball is thrown up, negative force of gravity is applied and so the speed becomes zero.After that due to gravitational force the ball falls down with same initial velocity.

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The figure below shows a man in a boat on a lake. The man's mass is 74 kg, and the boat's is 135 kg. The man and boat are initia
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The velocity of the boat after the package is thrown is 0.36 m/s.

<h3>Final velocity of the boat</h3>

Apply the principle of conservation of linear momentum;

Pi = Pf

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  • Pf is final momentum

v(74 + 135) = 15 x 5

v(209) = 75

v = 75/209

v = 0.36 m/s

Thus, the velocity of the boat after the package is thrown is 0.36 m/s.

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2 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
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Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

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And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

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