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Morgarella [4.7K]
3 years ago
10

What is it asking MathPhys im sorry i dont know but i tried teaching myself physics but i cant

Physics
1 answer:
RSB [31]3 years ago
6 0

Answer:

The puck B remains at the point of collision.

Explanation:

This is an elastic collision, so both momentum and energy are conserved.

The mass of both pucks is m.

The velocity of puck B before the collision is vb.

The velocity of puck A and B after the collision is va' and vb', respectively.

Momentum before = momentum after

m vb = m vb' + m va'

vb = vb' + va'

Energy before = energy after

½ m vb² = ½ m vb'² + ½ m va'²

vb² = vb'² + va'²

Substituting:

(vb' + va')² = vb'² + va'²

vb'² + 2 va' vb' + va'² = vb'² + va'²

2 va' vb' = 0

va' vb' = 0

We know that va' isn't 0, so:

vb' = 0

The puck B remains at the point of collision.

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3 years ago
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Answer:

Mass of the box = 0.9433 kg

Explanation:

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[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision v_{2i} = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision v_{2f}= 1.53 m/s

To find the mas of the box m_2.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

p_i=p_f

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

We can plugin the given value to find m_2

(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)

0.9522+0=-0.4911+1.53m_2

Adding both sides by 0.4911

0.9522+0.4911=-0.4911+0.4911+1.53m_2

1.4433=1.53m_2

Dividing both sides by 1.53.

\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}

0.9433=m_2

∴ m_2=0.9433 kg

Mass of the box = 0.9433 kg (Answer)

4 0
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Answer:

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Answer:

Explanation:

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