A 0.137 kg mass on a string
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When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
Explanation :
There are two types of collision i.e. elastic and elastic collision.
- Elastic collision : In this type of collision, the total momentum and the kinetic energy of the particles remains constant.
- Inelastic collision : In this type of collision, only the momentum remains constant while there is some loss of kinetic energy occurs.
From Newton's second law,
F = m a
a is the rate of change of velocity.
There is a inverse relation between the force and the time of collision.
The change in <em><u>momentum</u></em> will remain the same during a collision, the force needed to bring an object to a stop can be <em><u>increased</u></em> if the time of the collision is <u><em>decreased</em></u>.