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?

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sukljama2 · Accepted Answer

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  p→1=−2m2gh1  Step 3  Similarly, after the rebound, the momentum must be  p→2=+2m2gh2  Step 4  The impulse must be  I→=p→2−p→2  =+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 · Accepted Answer

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=mvf−mvi  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=mvf−mvi  =(0.150kg)(−4.34ms−4.95ms)  =−1.39kg×ms  The negative sign indicates that the direction of the impulse is upward.

nick1337 · Accepted Answer

Step 1 By definition, the impulse given to the ball is equal to the change of its momentum: J=Δp=m(v2−v1) where m=0.15kg is the mass of the ball,v1, v2 are the projections of the ball&#039;s velocity on the vertical axis before and after the collision respectively. Let&#039;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(v2−v1)=m2g(h1+h1) J=0.15×2×9.18×(1.25+0.96)≈1.39N×s

RizerMix · Accepted Answer

The initial potential energy of the ball is given by:PE1=mgh1where 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=mgh2where h2 is the rebound height (0.960 m).Since energy is conserved, the initial potential energy is equal to the final potential energy:mgh1=mgh2We can solve this equation for h2 to find:h2=h1g=1.259.8≈0.128mNow, we can calculate the velocity of the ball just before it hits the floor. The change in height is given by:Δh=h1−h2=1.25−0.128=1.122mUsing the equation for gravitational potential energy, we can write:ΔPE=mgh=12mv2where v is the velocity of the ball just before it hits the floor. We can solve this equation for v:v=2ghSubstituting the values for g and Δh, we get:v=2·9.8·1.122≈4.21m/sNow, 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·ΔvSince 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/s≈−1.263NsSo, the impulse given to the ball by the floor is approximately −1.263Ns (negative sign indicates a change in direction).

Vasquez · Accepted Answer

Result:−0.084NsSolution:Given:ΔPE=Work+ImpulseThe change in potential energy is given by:ΔPE=m·g·(h2−h1) 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)·(0.960m−1.25m)Simplifying the expression inside the parentheses:ΔPE=(0.150kg)·(9.8m/s2)·(−0.290m)Now, let&#039;s solve for the work done by the floor. The work done is given by the negative of the change in potential energy:Work=−ΔPESubstituting 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=ΔPE−WorkSubstituting 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:Impulse≈−0.084NsTherefore, the impulse given to the ball by the floor is approximately −0.084Ns.

A ball of mass 0.150 kg is dropped from rest

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