Using Hooke's law F spring = k dela x, find elastic constant of a spring that stretches 2cm when a 4 newton force is applied to
it
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Answer:
Mass of the box = 0.9433 kg
Explanation:
Mass of racket-ball = 0.00427 kg
Velocity of racket-ball before collision = 22.3 m/s
Velocity of racket-ball after collision with box = -11.5 m/s
[Since ball is bouncing back, so velocity is taken negative.]
Velocity of the box before collision = 0 m/s
<em>[Since the box is stationary, so velocity is taken zero]</em>
Velocity of box moving forward after collision = 1.53 m/s
To find the mas of the box .
By law of conservation of momentum we have:
Momentum before collision = Momentum after collision
This can be written as:
We can plugin the given value to find
Adding both sides by 0.4911
Dividing both sides by 1.53.
∴ kg
Mass of the box = 0.9433 kg (Answer)