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Alex17521 [72]
3 years ago
7

A 300-kg truck moving rightward with a speed of 5 km/hr collides head on with a 1000kg car moving leftward with a speed of 100 k

m/hr. The two vehicles stick together and move with the same velocity after the collision determine the post - collision speed of the car and truck
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

The final velocity of the vehicles will be -75km per hour (leftwards).

Explanation:

Let us call m_1, v_1 mass and velocity of the truck , m_2, v_2 mass and velocity of the car, and v_f the final velocity of the vehicles.

The law of conservation of momentum states that

m_1v_1-m_2v_2= v_f(m_1+m_2)

solving for v_f we get:

v_f = \dfrac{m_1v_1-m_2v_2}{m_1+m_2}

putting in numerical values

m_1 =300 kg \\v_1 =5km/hr = 1.39m/s\\m_2 = 1000kg\\v_2 = 100km/hr = 27.78m/s

v_f = \dfrac{300*(1.39)-1000*(27.78)}{300+1000}

\boxed{v_f = -21.0m/s } (<em>the negative sign indicates that the vehicles are moving in the leftward direction)</em>

or -75.8 km per hour.<em> </em>

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A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s
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There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

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Explanation:

Given data

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Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

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w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\   So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm

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