A ball of mass 0.5kg is dropped from a height of 10m and rebound with a velocity 1/3 of that before impact. The height reached after rebound is?
V^2 = Vo^2 + 2g*h = 0 + 19.6*10 = 196,
V = 14 m/s before impact.
Vr = V/3 = 14/3 = 4.67 m/s = Velocity at which the ball rebounds.
V^2 = Vr^2 + 2g*h = 0.
4.67^2 + (-19.6)h = 0,
h =
It's KEnergy is now (1/3)^2 the original (KE depends on velocity squared), so it can only go to1/3 the original height.
Well, well, well, looks like this ball is having quite the bounce party! Let's see what we can figure out here.
We know that the ball is dropped from a height of 10m, so the initial potential energy is given by mgh, where m is the mass (0.5kg), g is the acceleration due to gravity (approximately 9.8m/s²), and h is the height (10m). So, the initial potential energy is 0.5kg * 9.8m/s² * 10m, which gives us 49 Joules of potential energy.
Now comes the bouncy part. After the rebound, the ball has a velocity of 1/3 of what it had before impact. So, we can say that the final kinetic energy after the rebound is 1/9 times the initial kinetic energy before impact.
Since energy is conserved, the sum of the initial potential energy (49 Joules) and the initial kinetic energy (let's call it K) must be equal to the final kinetic energy after the rebound. So, we have:
49 Joules + K = 1/9K
Simplifying this equation, we find K = 441/8 Joules. This is the initial kinetic energy before impact.
Now, to find the height reached after the rebound, we calculate the potential energy at that point. Since the ball has achieved its maximum height, all its initial kinetic energy is converted to potential energy.
Using the equation mgh, where m is the mass (0.5kg), g is the acceleration due to gravity (again, approximately 9.8m/s²), and h is the height reached after the rebound, we can solve for h.
So, 441/8 Joules = 0.5kg * 9.8m/s² * h
Simplifying this equation, we find h = 441/0.4, which gives us a height of 1102.5 meters.
Wowza! That's one high bounce! I hope this explanation didn't bounce right over your head!
To find the height reached after rebound, we need to use the principle of conservation of mechanical energy.
The initial potential energy at a height of 10m is given by:
PE_initial = m * g * h
where m is the mass of the ball (0.5kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (10m).
The initial kinetic energy at the moment of impact is zero since the ball is dropped.
Therefore, the total mechanical energy (E_total) before impact is given by:
E_total = PE_initial + KE_initial
Since there is no initial kinetic energy, the mechanical energy before impact is equal to the potential energy:
E_total = PE_initial = m * g * h
After the ball rebounds, it reaches a new height. At this new height, the total mechanical energy is given by:
E'_total = PE_final + KE_final
The final potential energy (PE_final) at the new height is:
PE_final = m * g * h'
where h' is the rebound height that we need to find.
The final kinetic energy (KE_final) can be found using the velocity after rebound (v_rebound) which is one-third of the velocity before impact:
KE_final = (1/2) * m * v_rebound^2
Substituting the value of the velocity (v_rebound) into the formula, we have:
KE_final = (1/2) * m * (1/3 * v_before_impact)^2
Simplifying, we get:
KE_final = (1/2) * m * (1/9) * v_before_impact^2
The total mechanical energy after rebound (E'_total) is:
E'_total = PE_final + KE_final
We know that the total mechanical energy is conserved, so E_total = E'_total:
m * g * h = m * g * h' + (1/2) * m * (1/9) * v_before_impact^2
Canceling out the mass and acceleration due to gravity, we get:
h = h' + (1/18) * v_before_impact^2
Rearranging the equation to solve for h', we have:
h' = h - (1/18) * v_before_impact^2
Now we can substitute the given values into the equation.
Given:
m = 0.5kg
h = 10m
v_before_impact = ?
Since we are not given the velocity before impact, we cannot determine the rebound height without that information.
To find the height reached after the rebound, we need to use the principle of conservation of mechanical energy.
The total mechanical energy of the ball at any point can be given as the sum of its kinetic energy (KE) and potential energy (PE). Mathematically, it can be expressed as:
Total mechanical energy = KE + PE
When the ball is at the topmost point (before falling), it has only potential energy, given by:
PE = mgh
where:
m = mass of the ball (0.5 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the ball (10 m)
Thus, the initial potential energy of the ball is:
PE_initial = (0.5 kg) * (9.8 m/s^2) * (10 m) = 49 J
When the ball rebounds, it has both kinetic and potential energy (as it rises). Since we know the rebound velocity of the ball is 1/3 of its velocity before impact, we can write:
KE_rebound = (1/2) * m * (v_rebound^2)
To find the rebound velocity (v_rebound), we can use the principle of conservation of linear momentum. The linear momentum before impact is equal to the linear momentum after impact (assuming no external forces act on the ball). Mathematically, it can be expressed as:
m * v_initial = m * v_rebound
where:
v_initial = velocity before impact
Thus, v_rebound = v_initial / 3
Now, using the principle of conservation of mechanical energy (the sum of KE and PE remains constant), we can write the equation:
Total mechanical energy before = Total mechanical energy after
(KE_initial + PE_initial) = (KE_rebound + PE_rebound)
Since the ball is dropped from rest, its initial velocity is 0:
KE_initial = 0, PE_initial = 49 J
Substituting the known values:
49 J = (1/2) * (0.5 kg) * (v_rebound^2) + mgh_rebound
Substituting the value of v_rebound calculated earlier:
49 J = (1/2) * (0.5 kg) * ((v_initial/3)^2) + mgh_rebound
49 J = (1/2) * (0.5 kg) * ((v_initial^2)/9) + mgh_rebound
Simplifying:
h_rebound = [49 J - (1/2) * (0.5 kg) * ((v_initial^2)/9)] / [m * g]
Now, we can calculate the height reached after rebound by substituting the known values:
h_rebound = [49 J - (1/2) * (0.5 kg) * ((v_initial^2)/9)] / [(0.5 kg) * (9.8 m/s^2)]
Once the velocity before impact (v_initial) is known, we can calculate the height reached after rebound using this formula.