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Homework Statement
Consider a small body of mass m placed over a larger body of mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical at height h.
The smaller mass is pushed on the longer one at speed v and the system is left to itself.
Assuming all surfaces to be frictionless, find the speed of the smaller mass when it breaks off the larger mass.
My attempt
This problem has been frustrating me for quite some time now.
I start by assuming the bigger block is moving at speed V to the left when the smaller block is on the vertical and just about to break off.
Then by conserving momentum in horizontal direction : [tex] V=\frac{mv}{M+m} [/tex]
since the small block is also having a velocity component V to the left wrt ground
Now, i am going to use conservation of energy, and this is where I have doubts.
I am not sure whether to choose only the small block & ground as the system , or to choose both blocks & ground as the system. (I am getting answer in neither)
I think that in choosing only the small block and ground as system, an external force by the large block is acting (normal reaction) and thus i cannot use conservation of energy directly. Is this hypothesis correct ?
So, choosing the both blocks and ground as system :
The small block is having a velocity component upwards and velocity component left wards which is V. Let the required speed of the small block at the time of breaking off be x
Since [tex]F_{ext}=0[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + mgh + \frac{1}{2}mx^2 [/tex]
Plz verify whether this equation is correct
Solving, i get [tex]x^2=\frac{(v^2-2gh)(m+M)^2-mMv^2}{(m+M)^2} [/tex]
while the answer is printed as [tex]x^2=\frac{(M^2+Mm+m^2)}{(M+m)^2}v^2-2gh [/tex]
Sorry for the length, but i desperately need some input
Thx a lot !
Consider a small body of mass m placed over a larger body of mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical at height h.
The smaller mass is pushed on the longer one at speed v and the system is left to itself.
Assuming all surfaces to be frictionless, find the speed of the smaller mass when it breaks off the larger mass.
My attempt
This problem has been frustrating me for quite some time now.
I start by assuming the bigger block is moving at speed V to the left when the smaller block is on the vertical and just about to break off.
Then by conserving momentum in horizontal direction : [tex] V=\frac{mv}{M+m} [/tex]
since the small block is also having a velocity component V to the left wrt ground
Now, i am going to use conservation of energy, and this is where I have doubts.
I am not sure whether to choose only the small block & ground as the system , or to choose both blocks & ground as the system. (I am getting answer in neither)
I think that in choosing only the small block and ground as system, an external force by the large block is acting (normal reaction) and thus i cannot use conservation of energy directly. Is this hypothesis correct ?
So, choosing the both blocks and ground as system :
The small block is having a velocity component upwards and velocity component left wards which is V. Let the required speed of the small block at the time of breaking off be x
Since [tex]F_{ext}=0[/tex]
[tex]\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + mgh + \frac{1}{2}mx^2 [/tex]
Plz verify whether this equation is correct
Solving, i get [tex]x^2=\frac{(v^2-2gh)(m+M)^2-mMv^2}{(m+M)^2} [/tex]
while the answer is printed as [tex]x^2=\frac{(M^2+Mm+m^2)}{(M+m)^2}v^2-2gh [/tex]
Sorry for the length, but i desperately need some input
Thx a lot !