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dalvyx [7]
3 years ago
11

A green ball is dropped from rest from an elevated position at the same instant that a red ball is launched horizontally (from t

he same height), which ball will hit the ground first? Assume the balls behave as projectiles.
a. green ball
b. red ball
c. the red and green ball will hit the ground at the same time
d. not enough information given
Physics
1 answer:
schepotkina [342]3 years ago
6 0

Answer:

c. the red and green ball will hit the ground at the same time

Explanation:

if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

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A 0.0427 kg racquet-ball is moving
Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball (m_1) = 0.00427 kg

Velocity of racket-ball before collision (v_{1i}) = 22.3 m/s

Velocity of racket-ball after collision with box (v_{1f}) = -11.5 m/s

[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)

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