Question

Can I solved this problem by using conservation of energy ?

A bullet of mass m moving with a horizontal velocity u strikes a stationary wooden block of mass M suspended by a string of length L=50 cm.The bullet emerges out the block with the speed u/4.If M=6m, the minimum value of u so that the block can complete the vertical circle?

Answer 1

I got `11ms^-1` but check my maths anyway...

Explanation:

Have a look:

Answer 2

`35.42 m/s`

Explanation:

`"Momentum before collision " = " Momentum after collision"`

`m_1u_1+m_2u_2=m_1v_1+m_2v_2` ----- (1)

For Bullet

• `m_1=m`
• `u_1=u`
• `v_1=u/4`

For Wooden Block

• `m_2=6m`
• `u_2=0`
• `v_2=color(red)?`

From (1),

`m u+0=m u/4+6mcolor(red)(v_2)` ----- (2)

Now, to find `color(red)(v_2)` of wood block, [Refer the image]

`"Displacement"(S) = 0.5*2=1m`
`"acceleration"(a)=-9.8m/s^2`
`"final velocity"(v_f)=0m/s("its velocity is zero at the top")`
`"initial velocity"(color(red)(v_2))=?`

Now, We know that `2aS=v^2-u^2` (kinematic equation `"III"`)

So, `2(-9.8)(1)=0^2-z^2`
`=>color(red)(v_2)^2=19.6`
`=> color(red)(v_2)= sqrt(19.6)`

Now, substituting it in (2) `->`

`m u=m u/4+6m*sqrt(19.6)`

Simplifying, we get,
`u=35.42m/s`

Answer 3

`12.52 ms^-1`

Explanation:

For a particle of mass `M` to complete its motion in a vertical circle,at the highest point of the motion,the string tension should be just zero,so that the necessary centripetal force for the movement along a circle for that moment is supplied by the weight only,

So,we can write,if velocity at highest point is `v` then, `(mv^2)/r =mg`

or, `v^2 =gr`

So,applying law of conservation of energy,we can find what must be the velocity(`V`) at the bottom point of this vertical circle,

So, `mgh+ 1/2 mv^2 =1/2 mV^2`

So, `mg(2r) + 1/2 mgr = 1/2 m V^2`

or, `V^2 =5gr`

Now,the bullet must hit the block in such a fashion,that this mass `M` can get a velocity of `sqrt(5gr)` after the collision.

So,applying law of conservation of energy during collision,we can say (considering no form of energy loss)

`1/2 m u^2 +0 = 1/2 m (u/4)^2 + 1/2 m 5gr`

Given, `r=50/100=1/2 m`

So,solving we get, `u=12.52 ms^-1`

Answer 4

This is what I obtain.

Explanation:

First step. Use Law of Conservation of momentum to calculate velocity `v_w` of the wooden block just after the collision.

Momentum before collision `=` Momentum after collision

`m u+0=m u/4+6mv_w`
`=> u=8v_w` ............ (1)

Next step. To find minimum value of `v_(w_min)` so that the block can complete the vertical circle.

Kinetic energy imparted to the block will get converted to GPE of the block at the top of the circle and its velocity will be zero.

`1/2(6m)v_(w_min)^2=(6m)(10)(1)`
`=>v_(w_min)^2=20`
`=>v_(w_min)=sqrt20` ........(2)

Step 3. To confirm if this velocity is sufficient for the motion of wooden block in a vertical circle.

`v_(w_min)=sqrt(5gr)`

Inserting given values we get

`v_(w_min)=sqrt(5xx10xx0.5)`
`v_(w_min)=sqrt(25)`
`v_(w_min)=5` .........(3)

We need to have minimum velocity as higher of the two in (2) and (3)

Inserting in (1)

`u=8xx5`
`u=40\ ms^-1`

.-.-.-.-.-.-

We actually need not complete step 2.
As we know that in this step we get `v=sqrt(4gr)`
This value is always `"<"sqrt(5gr)` the value we are going to calculate in step 3.

Cheers