Thushari Jaya sekera 1. Question 1 is composed of 4 aparts. Answer each part in

Question

Thushari Jaya sekera 1. Question 1 is composed of 4 aparts. Answer each part in detail and show all work. (a) 7 points A 6.0kg object moving 5.0m/s collides with and sticks to a 2.0kg object. After the collision the composite object is moving 2.0m/s in a direction opposite to the initial direction of motion of the 6.0kg object. Determine the speed of the 2.0kg object before the collision. (b) 7 points A 2.0kg object moving 3.0m/s strikes a 1.0kg object initially at rest. Immedi ately after the co on, the 2.0kg object has a velocity of 1.5m/s directed 300 from its initial direction of motion. What is the i component of the velocity of the 1.0kg object just after the collision?

Explanation / Answer

1.(a)

Let the direction of the intitial velocity of the 6 kg block be positive. Let the speed of 2 kg block be v before collision.

Now applying conservation of momentum principle,we have

(6 kg)(5 m/s) + (2 kg)v = -(6 kg + 2kg)(2 m/s)

or, v = -23 m/s

So, before collision, the 2 kg block is moving at a speed of 23 m/s opposite to the 6 kg block.

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1.(b)

Applying momentum conservation along x-axis, we have

(2 kg)(3 m/s) + 0 = (2 kg)(1.5 m/s)cos300 + (1 kg)(vx)

or, vx = 3.4 m/s

So, the x-component of velocity of 1 kg block after collision is 3.4 m/s

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1.(c)

Impulse is equal to the area under F-t curve.

So impulse P = (1/2)(18000 N)(1.5ms) = (0.5)(18000 N)(1.5X10-3s)

or, P = 13.5 N-s

Average force exerted is,

F = 13.5N-s/1.5ms

F = 9000N

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1.(d)

Initial speed is u = 6 m/s

final speed v = 0

let he ascends a height h

So, using the formula, v2 - u2 = 2as, where v= 0, g = -9.8 m/s2

we get, 0 - (6 m/s)2 = (2)(-9.8 m/s2)h

or, h = 1.84 m

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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