Mechanics> A block of mass M is pushed towards a mov...

A block of mass M is pushed towards a movable wedge of mass k M and height H with a velocity U all surfaces are smooth the minimum value of u for which the block will reach the top of the wedge isa)√2ghb)k√2ghc)√2gh(1+1/k)

Rock , 6 Years ago

Grade 12

2 Answers

Arun

Last Activity: 6 Years ago

Now that every surface is smooth, no loss of energy will be observed.
Therefore the Minimum condition would be for the mass to climb the greatest height and come to rest, meaning at the top of the wedge and come to rest wrt the wedge. But the wedge would already have started moving because of the Impact. Therefore at the end, they move as a single entity with some velocity 'v', say.
Now conserve the Momentum,
mu = (km+m)*v....(1)
Conserve the energy,
0.5*m*u^2 = 0.5*(km+m)*v^2 +mgh....(2)
Solve both the equations (by eliminating 'v') and we can have the Minimum value of 'u'.

Answer will come as option C

ankit singh

Last Activity: 4 Years ago

Let the mass of block is m

Mass of wedge is M

Initial velocity of block is u

Height of wedge is h

As we want the minimum value of u , for that the block must have a final potential energy of mgh and it's velocity only in horizontal direction , the final velocity (v) of both, the block and the wedge must be the same in the horizontal direction to prevent falling back of the block.

Now by conserving total energy of the system.

Initial energy = final energy

~0.5 muu = 0.5( M+m)vv + mgh

~uu = [(M+m)/(m)]vv + 2gh

Now by conserving momentum in horizontal direction

mu = (m+M)v

v = (mu)/(m+M)

By substituting the value of v from the later eqation into the previous equation. We get,

uu = (mu)/(m+M) + 2gh

By solving this quadratic equation, there will be two values of u , one will be positive and other will be negative.

I am not solving the equation. It is very difficult to type it down.