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```
A shell is fired from a canon with a velocity v(m/sec) at an angle x with the horizontal.At the highest point in its path, it explodes into 2 pieces of equal masses.One of the piece retraces into the cannon then the speed in m/sec of the other piece immediately after the explosion is; (a)3v cos x (b)2v cos x (c)1.5v cos x (d)1.5^2v cos x

```
11 years ago

```							Hi
ans: 3vcosx
At the top of its path, vel is vcos x only in horizontal direction , so after explosion to retrace its path , the velocity of one of the equal masses should be -vcos x . therefore now if v conserve the momentum, then 2mvcosx = -mvcosx + mv' . calculate v' .
```
11 years ago
```							hi vaibhav,
dis question is quite straight forward
as velocity at d highest point is v cosx
just apply momentum conservation
m(v cosx) = m/2 (-ucosx ) + m/2 v1,   [ as it is divided into parts of equal mass n one part retraces to d cannon itself so its velocity will be vcosx but wid a negative sign]
we hav to find v1
here simply u get v1= 3v cosx
```
11 years ago
```							Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘?’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cos?. This component remains constant.
At the maximum height, the cannon has only this component as the vertical component is momentarily zero.

So, momentum before exploding = 2mv cos?
After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cos?’.
Therefore, momentum after explosion = -mv cos? + mu
So, by conservation of momentum,
2mv cos? = -mv cos? + mu
= > u = (3v cos?) m/s
```
6 years ago
```							Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘?’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cos?. This component remains constant.At the maximum height, the cannon has only this component as the vertical component is momentarily zero.So, momentum before exploding = 2mv cos?After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cos?’.Therefore, momentum after explosion = -mv cos? + muSo, by conservation of momentum,2mv cos? = -mv cos? + mu= > u = (3v cos?) m/s
```
4 years ago
```							Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘?’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cos?. This component remains constant.At the maximum height, the cannon has only this component as the vertical component is momentarily zero.So, momentum before exploding = 2mv cos©After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cos?’.Therefore, momentum after explosion = -mv cos? + muSo, by conservation of momentum,2mv cos? = -mv cos© + mu= > u = (3v cos©)m/s
```
4 years ago
```							Dear friends,By conservation of momentumInitial momentum = Final momentumInitially the mass of Canon is 2m and its horizontal component is vcosxInitially mass of Canon is taken because at that time the bullet is not explode into two partsBut the final momentum of bullet is break down into two parts the two particles which explodes both have same mass but have different velocities.The one which retraced is path have velocity -vcosx and we have to find velocity of second particleInitial momentum=Final momentum2mvcosx = -mvcosx + mumu = 2mvcosx + mvcosxmu = m(2vcosx + vcosx)u = 3vcosx(1) ........
```
4 years ago 605 Points
```							Dear student,Please find the solution to your solution to your problem below. At the top of its path, vel is vcos x only in horizontal direction , so after explosion to retrace its path , the velocity of one of the equal masses should be -vcos x . therefore now if we conserve the momentum, then 2mvcosx = -mvcosx + mvTherefore, v = 3vcosx Hope it helps,Thanks and regards,Kushagra
```
4 months ago
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