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Arturiano [62]
3 years ago
13

A 0.50 Kg billard ball moving at 1.5 m/s strikes a second 0.50 kg billiard ball which is at rest on the table.If the first ball

slows down to a speed of 0.10 m/s, then what is the speed of the second ball ? Please explain.
Physics
1 answer:
vichka [17]3 years ago
6 0

Answer:

velocity of second billiard after collision = 1.4 m/sec  

Explanation:

We have mass of billiard let m_1=0.5kg

Velocity of billiard let v_1=1.5m/sec

Mass of second second billiard let m_2=0.5kg

Velocity of second billiard before collision 0 m/sec  billiard

Velocity of first billiard after collision  0.1 m/sec

Now according conservation of momentum

Momentum before collision and after collision will be same

So momentum before collision = momentum after collision

0.5\times 1.5+0.5\times 0=0.5\times 0.1+0.5\times velovity\ of\ second\ biliard\ after\ collision

So velocity of second billiard after collision = 1.4 m/sec  

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

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If the direction of the position is north and the direction of the velocity is up, then what is the direction of the angular mom
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Answer:

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