Newton's laws--Block sliding up a frictionless semi-circular track

[Dale]

just want to know if the x velocities are not the same in the curve

They are not, except at point A.

how the bodies dnt loose contact or what velocities are equal for the contact to be maintained

The component of the velocities in the normal direction must be equal to maintain contact

So is everything resolved? If not, what remains? Please be as clear as possible

 

[jbriggs444]

I just want to know if the x velocities are not the same in the curve

It is far from clear. But I think that you are confusing velocity with speed, invoking conservation of momentum, adopting a center of momentum frame and forgetting that the two blocks may have different masses.

If the two blocks have equal masses and start with zero total momentum then conservation of momentum immediately demands that the horizontal component of their velocities are at all times equal and opposite so that their horizontal speeds are at all times equal.

 

[sachin]

if the x velocities are different,why does not the small block pierce through and enter inside the bigger block.

 

[Dale]

if the x velocities are different,why does not the small block pierce through and enter inside the bigger block.

Because the direction “inside the bigger block” is the normal direction, not the x direction

 

[erobz]

W.r.t the ground the velocity of the sled when the mass is at some angle $\theta$ is represented by the green arrow

W.r.t the ground the velocity of the mass on the sled in the horizontal direction is the projection of the red arrow onto the horizontal direction plus the green arrow.

In general, the velocity of the sled in the horizontal direction will not equal the velocity of the mass in the horizontal direction.

It is at point A, when the projection of the red arrow onto the green arrow is $0$, that the velocity of the sled matches the velocity of the mass.

 

[sachin]

View attachment 312968

W.r.t the ground the velocity of the sled when the mass is at some angle $\theta$ is represented by the green arrow

W.r.t the ground the velocity of the mass on the sled in the horizontal direction is the projection of the red arrow onto the horizontal direction plus the green arrow.

In general, the velocity of the sled in the horizontal direction will not equal the velocity of the mass in the horizontal direction.

It is at point A, when the projection of the red arrow onto the green arrow is $0$, that the velocity of the sled matches the velocity of the mass.

yes,its all true,the red arrow has only horizontal and vertical components,has not got any component along the common normal i.e perpendicular to the red one,for the bodies to be in contact the velocities along the common normal should be the same,how the contact still remains ?

 

[erobz]

yes,its all true,the red arrow has only horizontal and vertical components,has not got any component along the common normal i.e perpendicular to the red one,for the bodies to be in contact the velocities along the common normal should be the same,how the contact still remains ?

They are the same. The velocities of both blocks along the common normal is $0$. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.

 

[sachin]

They are the same. The velocities of both blocks along the common normal is $0$. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.
yes,its quite clear now,thanks a lot.
View attachment 312996

They are the same. The velocities of both blocks along the common normal is $0$. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.

View attachment 312996

yes,its all clear,hope the same relationship holds true for displacement and acceleration also,thanks.