A ball of mass 0.150 kg is dropped from rest

Sam Longoria

Sam Longoria

Answered question

2021-12-18

A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. What impulse was given to the ball by the floor?

Answer & Explanation

sukljama2

sukljama2

Beginner2021-12-19Added 32 answers

Step 1
In this problem, a ball of mass m=0.150m is dropped from a rest height of h1=1.25m. If rebounds from the floor to reach of h2=0.960m. We calculate the impulse given to the ball by the floor. We use g=9.80ms2
Step 2
Let the upward direction be positive. Initially, the ball moves downwards, so the initial momentum is negative. Using conservation of energy, we have
KEi+PEi=KEf+PEf
0+mgh1=mv122+0
mgh1=m2v122m
mgh1=p122m
p12=2m2gh1
p1=2m2gh1
p1=2m2gh1
Step 3
Similarly, after the rebound, the momentum must be
p2=+2m2gh2
Step 4
The impulse must be
I=p2p2
=+2m2gh2(2m2gh1)
=2m2gh2+2m2gh1
=2m2g(h2+h1)
=[2(0.150kg)2(9.80ms2)](0.960m+1.25m)
=+1.39312N×s
I=+1.39N×s
Since the impulse is positive, it must be upwards.
Becky Harrison

Becky Harrison

Beginner2021-12-20Added 40 answers

When an object receives a force for a short time, this force with time is called the impulse, so it is the area under the curve of the force-versus-time graph and it is the same as the change in momentum. The impulse is the quantity I and it is given by equation (6.4) in the form
1) I=FΔt=mvfmvi
When the ball is dropped from a height of 1.25 m, it gains velocity due to the gravitational acceleration. This is the initial velocity before it hits the floor. From the kinematic equations, we can calculate this velocity by
2) vi2=v02+2gh0
vi=0+2gh0
The height is h0=1.25m Use this value into equation (2) to get the velocity of the ball before it drops on the floor
vi=2gh0=2(9.8ms2)(1.25m)=4.95ms
When the ball hits the floor, it changes its velocity. The final velocity could be calculated using equation (2) but for height, h=0.960m
vf=2gh=2(9.8ms2)(0.960m)=4.34ms
Now, we plug the values for m, vf and vi into equation (1)to get the impulse
I=mvfmvi
=(0.150kg)(4.34ms4.95ms)
=1.39kg×ms
The negative sign indicates that the direction of the impulse is upward.
nick1337

nick1337

Expert2021-12-27Added 777 answers

Step 1
By definition, the impulse given to the ball is equal to the change of its momentum:
J=Δp=m(v2v1)
where m=0.15kg is the mass of the ball,v1, v2 are the projections of the ball's velocity on the vertical axis before and after the collision respectively.
Let's direct the vertical axis upwnward.
According to the mechanical energy conservation, the kinetic energy of the ball right before the collision is equal to its potential energy at the begining of the motion. Writting down the corresponding formulas for these energies, obtian:
mv122=mgh1
where g=9.81m/s2 is the gravitational acceleration. and h1=1.25m is the initial height. Thus, obtain:
v1=2gh1
(negative, since before the collision the velocity is directed downward).
After the collision, the kinetic energy of the ball is equal to its potential energy at the maximum height h2=0.96m. Obtain:
mv222=mgh2
v2=2gh2
(positive, since after the collision the velocity is directed upward).
Finally, obtain the expression for the impulse:
J=m(v2v1)=m2g(h1+h1)
J=0.15×2×9.18×(1.25+0.96)1.39N×s

RizerMix

RizerMix

Expert2023-06-18Added 656 answers

The initial potential energy of the ball is given by:
PE1=mgh1
where m is the mass of the ball (0.150 kg), g is the acceleration due to gravity (9.8 m/s^2), and h1 is the initial height (1.25 m).
The final potential energy of the ball at height h2 is given by:
PE2=mgh2
where h2 is the rebound height (0.960 m).
Since energy is conserved, the initial potential energy is equal to the final potential energy:
mgh1=mgh2
We can solve this equation for h2 to find:
h2=h1g=1.259.80.128m
Now, we can calculate the velocity of the ball just before it hits the floor. The change in height is given by:
Δh=h1h2=1.250.128=1.122m
Using the equation for gravitational potential energy, we can write:
ΔPE=mgh=12mv2
where v is the velocity of the ball just before it hits the floor. We can solve this equation for v:
v=2gh
Substituting the values for g and Δh, we get:
v=2·9.8·1.1224.21m/s
Now, we can calculate the impulse given to the ball by the floor. Impulse is defined as the change in momentum, and it can be calculated using the formula:
Impulse=m·Δv
Since the ball rebounds, its velocity changes from v to v. Therefore, the change in velocity, Δv, is 2v. Substituting the values, we have:
Impulse=0.150kg·(2·4.21)m/s1.263Ns
So, the impulse given to the ball by the floor is approximately 1.263Ns (negative sign indicates a change in direction).
Vasquez

Vasquez

Expert2023-06-18Added 669 answers

Result:
0.084Ns
Solution:
Given:
ΔPE=Work+Impulse
The change in potential energy is given by:
ΔPE=m·g·(h2h1) where m=0.150kg is the mass of the ball and g=9.8m/s2 is the acceleration due to gravity.
Substituting the given values, we have:
ΔPE=(0.150kg)·(9.8m/s2)·(0.960m1.25m)
Simplifying the expression inside the parentheses:
ΔPE=(0.150kg)·(9.8m/s2)·(0.290m)
Now, let's solve for the work done by the floor. The work done is given by the negative of the change in potential energy:
Work=ΔPE
Substituting the numerical values:
Work=(0.150kg)·(9.8m/s2)·(0.290m)
Finally, we can find the impulse given to the ball by rearranging the equation:
Impulse=ΔPEWork
Substituting the numerical values:
Impulse=(0.150kg)·(9.8m/s2)·(0.290m)(0.150kg)·(9.8m/s2)·(0.290m)
Simplifying further:
Impulse=2·(0.150kg)·(9.8m/s2)·(0.290m)
Calculating the numerical value, we find:
Impulse0.084Ns
Therefore, the impulse given to the ball by the floor is approximately 0.084Ns.

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