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torth76
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Here are 3 problems I have been working on. Any help would be very much appreciated. Thank you in advance!
Two inelastic balls are traveling toward each other with velocities -4m/s and 7 m/s and they experience an elastic head on collision
a)obtain the velocities (magnitude and direction) of each ball after the collision
b) recalculate the results for the ball moving at 7m/s being 2x the mass of the other ball
m1v1 + m2v2 = m1V1 + m2V2
v1 + V1 + v2 + V2
where v= initial and V=final
i feel to solve this i need to substitute one of the equations into the other and solve for one of the final velocities.
I did this and obtained
m1v1 + m2v2 - m1v2 + m1v1 = V (m1 +m2)
since masses are equal i factored like this
m(v1 +v2 -v2 +v1)/2m = V
m(v1+v1)/2m = V
mv1(2)/2m = V
v1=V
I feel like this is wrong. I have done it a few times and get different answers. I don't account for the direction of ball 2 until i plug in a value as -4, until then I leave it as positive.
A 0.5kg mass of silly putty is at rest at the top of a 100m tall building; it is dropped. At the same time a 0.003kg paintball pellet is fight straight up towards the ground with an initial veloctiy of 150 m/s. The paintball and putty collide inelastically 2.18m below the top of the building. Measured from the ground what is the maximun height that the combined putty/paintball mass achieves?
v^2 = v0^2 +2ax
m1v01 + m2v02 = (m1 +m2)Vf
i first used kinematics to find the velocities of the objects right at impact. the putty = -6.54m/s and the pellet =143.46.
Next I used m1v01 + m2v02 = (m1 +m2)Vf to solve for the velocity of the combined mass right after the collision. I found it to be -5.65m/s. I am not sure why my answer is negative and can't see why but I know it should be a positive value. If i use the positive value and use kinematics to solve for the height once it decelerates to v=0m/s i find it goes 1.63m equaling 99.45m above the ground.
I am unsure of this answer. I feel like my method is right but I am not sure if what I did was correct.
1 0.5kg mass of silly putty in suspended 5m above the floor by a long string. A paint ball pellet (m=0.003kg) is fired at the putty from the floor with an initial velocity of 150m/s. the two masses collide inelastically. what is the minimum length of the string such that the combined putty/paint ball mass does not splatter into the ceiling?
v^2 = v0^2 + 2ax
m1v01 +m2v02 = )m1+m2)Vf
using kinematics i determined the speed of the pellet before impact would be 149.67m/s. Using the equation for inelastic collision i found that the v0 of the ball is 0 and the vf of the combined mass to be 0.89m/s. Using this value I used the kinematics equation again and found distance (x) it would go until it came to rest. I found this value to be 0.04m.
After working through this problem I don't see what I am doing wrong. I don't think my answer makes sense since the string would only have to be longer than 4cm.
Homework Statement
Two inelastic balls are traveling toward each other with velocities -4m/s and 7 m/s and they experience an elastic head on collision
a)obtain the velocities (magnitude and direction) of each ball after the collision
b) recalculate the results for the ball moving at 7m/s being 2x the mass of the other ball
Homework Equations
m1v1 + m2v2 = m1V1 + m2V2
v1 + V1 + v2 + V2
where v= initial and V=final
The Attempt at a Solution
i feel to solve this i need to substitute one of the equations into the other and solve for one of the final velocities.
I did this and obtained
m1v1 + m2v2 - m1v2 + m1v1 = V (m1 +m2)
since masses are equal i factored like this
m(v1 +v2 -v2 +v1)/2m = V
m(v1+v1)/2m = V
mv1(2)/2m = V
v1=V
I feel like this is wrong. I have done it a few times and get different answers. I don't account for the direction of ball 2 until i plug in a value as -4, until then I leave it as positive.
Homework Statement
A 0.5kg mass of silly putty is at rest at the top of a 100m tall building; it is dropped. At the same time a 0.003kg paintball pellet is fight straight up towards the ground with an initial veloctiy of 150 m/s. The paintball and putty collide inelastically 2.18m below the top of the building. Measured from the ground what is the maximun height that the combined putty/paintball mass achieves?
Homework Equations
v^2 = v0^2 +2ax
m1v01 + m2v02 = (m1 +m2)Vf
The Attempt at a Solution
i first used kinematics to find the velocities of the objects right at impact. the putty = -6.54m/s and the pellet =143.46.
Next I used m1v01 + m2v02 = (m1 +m2)Vf to solve for the velocity of the combined mass right after the collision. I found it to be -5.65m/s. I am not sure why my answer is negative and can't see why but I know it should be a positive value. If i use the positive value and use kinematics to solve for the height once it decelerates to v=0m/s i find it goes 1.63m equaling 99.45m above the ground.
I am unsure of this answer. I feel like my method is right but I am not sure if what I did was correct.
Homework Statement
1 0.5kg mass of silly putty in suspended 5m above the floor by a long string. A paint ball pellet (m=0.003kg) is fired at the putty from the floor with an initial velocity of 150m/s. the two masses collide inelastically. what is the minimum length of the string such that the combined putty/paint ball mass does not splatter into the ceiling?
Homework Equations
v^2 = v0^2 + 2ax
m1v01 +m2v02 = )m1+m2)Vf
The Attempt at a Solution
using kinematics i determined the speed of the pellet before impact would be 149.67m/s. Using the equation for inelastic collision i found that the v0 of the ball is 0 and the vf of the combined mass to be 0.89m/s. Using this value I used the kinematics equation again and found distance (x) it would go until it came to rest. I found this value to be 0.04m.
After working through this problem I don't see what I am doing wrong. I don't think my answer makes sense since the string would only have to be longer than 4cm.