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Leviafan [203]
3 years ago
7

A 92-kg rugby player running at 7.5 m/s collides in midair with a 112-kg player moving in the opposite direction. After the coll

ision each player has zero velocity.
What was the initial speed of the 112-kg player before the collision?
Physics
1 answer:
madam [21]3 years ago
8 0

By the condition of momentum conservation we can say that since there is no external force along horizontal direction so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

here we know that

m_1 = 92 kg

v_{1i} = 7.5 m/s

m_2 = 112 kg

also we know that

v_{1f} = v_{2f} = 0

now from above equation we have

92(7.5) + 112 v = 0 + 0

now we have

v = 6.16 m/s

so the speed of the other player must be 6.16 m/s

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A mass m is attached to an ideal massless spring. When this system is set in motion, it has a period t. What is the period if th
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If a mass m is attached to an ideal massless spring and has a period of t, then the period of the system when the mass is 2m is \sqrt{2}t.

Calculation:

Step-1:

It is given that a mass m is attached to an ideal massless spring and the period of the system is t. It is required to find the period when the mass is doubled.

The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T.

It is known that the period is calculated as,

T=2 \pi \sqrt{\frac{m}{k}}

Here m is the mass of the object, and k is the spring constant.

Step-2:

Thus the period of the system with the first mass is,

t=2 \pi \sqrt{\frac{m}{k}}

The period of the system with the second mass is,

\begin{aligned}\\t^'&=2 \pi \sqrt{\frac{m}{k}}\\&=\sqrt{2}\times2 \pi \sqrt{\frac{2m}{k}}\\&=\sqrt{2}\times t\end{aligned}

Then the period of the system with the second mass is \sqrt{2} times more than the period of the system with the first mass.

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The tension in a horizontal spring is directly proportional to the extension (1 mark)
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This question involves the concepts of work done and elastic potential energy.

The work done by the spring when its extension changes from X to X/4 is "0.56 E".

The work done by the spring is equal to the elastic potential energy stored in the spring, which is given as follows:

W=E\\W=E=\frac{1}{2}kX^2---- eqn(1)

where,

W = work done = ?

E = elastic potential energy

k = spring constant

X = extension

Now, the spring moves to an extension of X/4, so the change in extension will be:

\Delta X = X-\frac{X}{4}\\\\\Delta X = \frac{3X}{4}

Hence, the work done will become:

W=\frac{1}{2}K\Delta X^2\\\\W=\frac{1}{2}K(\frac{3X}{4})^2\\\\W=\frac{1}{2}KX^2(0.56)\\\\using\ eqn(1):

<u>W = 0.56 E</u>

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Answer:

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