an object of mass 2 kg is dropped from a certain height.On rebounding from the ground,it rises vertically till 2/5th of its initial height.what will be the ratio of magnitude of momentum of the object just before and after striking the ground?
To solve this problem, we'll break it down into steps:
Step 1: Find the initial potential energy (PE) and momentum (P) of the object before it hits the ground.
Given:
Mass of the object, m = 2 kg
Acceleration due to gravity, g = 9.8 m/s²
Initial height, h = ?
Final height after rebounding, h' = 2/5 * h (Given)
Potential energy before hitting the ground is given by:
PE = m * g * h
Momentum before hitting the ground is given by:
P = m * √(2 * g * h)
Step 2: Find the potential energy (PE') and momentum (P') after the object rebounds.
Potential energy after rebounding is given by:
PE' = m * g * (2/5 * h)
Momentum after rebounding is given by:
P' = m * √(2 * g * (2/5 * h))
Step 3: Find the ratio of the magnitudes of momentum before and after striking the ground.
Ratio = P / P'
Now let's substitute the given values and solve the problem:
Initial potential energy:
PE = 2 kg * 9.8 m/s² * h
Initial momentum:
P = 2 kg * √(2 * 9.8 m/s² * h)
Final Potential energy after rebounding:
PE' = 2 kg * 9.8 m/s² * (2/5 * h)
Final Momentum after rebounding:
P' = 2 kg * √(2 * 9.8 m/s² * (2/5 * h))
Now we can calculate the ratio:
Ratio = P / P'
= (2 kg * √(2 * 9.8 m/s² * h)) / (2 kg * √(2 * 9.8 m/s² * (2/5 * h)))
Simplifying this expression will give you the final answer.
To find the ratio of the magnitude of momentum just before and after striking the ground, we need to consider the conservation of momentum.
First, let's break down the problem and find the initial and final velocities of the object.
Given:
- Mass of the object (m) = 2 kg
- The object is dropped from a certain height, which implies it begins its motion with a velocity of zero.
When the object rebounds from the ground:
- It rises to a height that is 2/5 of the initial height, which means it reaches 2/5 of its initial potential energy.
- The object reaches the maximum height and then falls back towards the ground.
Now, let's find the initial and final velocities.
Initial velocity (u): Since the object is dropped, the initial velocity is zero.
Final velocity (v): We can find the final velocity using the conservation of mechanical energy. The object reaches 2/5 of its initial height, which corresponds to 2/5 of its initial potential energy. At this point, all of its initial potential energy is converted into kinetic energy.
Using the conservation of mechanical energy:
Initial potential energy = Final kinetic energy
mgh = (1/2) mv^2
Here, h represents the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and v is the final velocity.
Since the initial velocity (u) is zero, the initial kinetic energy is also zero.
Thus, we have:
0 = (1/2) mv^2
0 = (1/2) × 2 kg × v^2
0 = v^2
Therefore, the final velocity (v) is zero when the object reaches 2/5 of its initial height.
Now let's calculate the ratios of momentum.
Just before striking the ground:
Momentum (p) = mass (m) × velocity (u)
Since the object is dropped, the initial velocity (u) is zero.
Therefore, the magnitude of momentum just before striking the ground is:
p_before = m × u
= 2 kg × 0
= 0 kg·m/s
Just after striking the ground:
The final velocity (v) is zero.
Therefore, the magnitude of momentum just after striking the ground is:
p_after = m × v
= 2 kg × 0
= 0 kg·m/s
Now, to find the ratio of the magnitudes of momentum just before and after striking the ground:
p_before / p_after = 0 kg·m/s / 0 kg·m/s
The ratio gives an undefined value since both numerator and denominator are zero.
Hence, the ratio of the magnitudes of momentum just before and after striking the ground is undefined.