### Discussion :: Engineering Mechanics - Section 1 (Q.No.3)

Divya said: (Nov 24, 2012) | |

Conservation of momentom mv=m1v1+m2v2 mv=mv`+mv` v`=v/2 |

Shrikant said: (Feb 6, 2013) | |

Two balls of equal mass and of perfectly elastic material are lying on the floor. One of the ball with velocity v is made to struck the second ball. Both the balls after impact will move with a velocity. |

Kumar Srinu said: (Feb 8, 2013) | |

Perfectly elastic materials e (coefficient of restitution) =1. |

Shiva Rao said: (Mar 16, 2013) | |

Velocity distributed uniformly. |

Saurabh said: (May 4, 2013) | |

One of the body which was moving initially with velocity V would come to rest and second body will move with velocity v. |

Raghavendra said: (Jun 8, 2013) | |

The velocity will be transferred from the 1st body to the second. |

Aditya said: (Jun 11, 2013) | |

The body transfer the whole energy to second body because e=1. So first body comes to rest. Another one moves velocity v. |

Pabitra said: (Jul 21, 2013) | |

Total velocity =v. Velocity distributed uniformly that = v. 2 ball each has v/2. |

Manohar said: (Aug 5, 2013) | |

Total momentum before and after collision is constant. Hence, m(v+0) = m(v'+v'). mv = 2mv'. v' = v/2. |

Ramana said: (Aug 7, 2013) | |

Two balls are same mass and same nature with the contact motion so each one have half of the original velocity. |

Gaurav Maini said: (Aug 7, 2013) | |

You are right guyz. It will follow conservation of momentum. But if it is perfectly elastic then the 1st ball should come at rest after colliding with the 2nd one. That means final velocity of 2nd ball will change from 0 to V. How it is v/2? Maybe I'm missing something. Please correct me guyz. |

Sandeep said: (Aug 31, 2013) | |

After hitting both transfer there energy to move v/2. |

Abhishek said: (Sep 8, 2013) | |

The first ball which was moving with velocity will become in rest and this velocity is transferred to the second ball and it will start moving with velocity we. So, none of the option is correct. |

Vinoth said: (Sep 14, 2013) | |

Total mass = mv'+mv'. mv = mv'+mv'. v = 2v'. And:v' = v/2. |

Manish Saini said: (Sep 17, 2013) | |

Total mass=mv'+mv' therefore v=2v' and v'=v/2. |

Tom said: (Oct 27, 2013) | |

The first ball will be at rest and the second will have a velocity v unless the collision angle is not zero, then in a certain angle of collision the two balls will have the same velocity. |

Sandeep said: (Dec 29, 2013) | |

There is no energy loss in perfectly elastic collision and hence momentum will be conserved. |

Vikas.G said: (Jan 18, 2014) | |

Velocity of first moving ball is v after impact its velocity becomes half therefor every action the is equal and opposite reaction so velocity of 2nd ball is also v/2. |

K. Srinivasan said: (Jan 23, 2014) | |

We know, Momentum before collision is equal to momentum after collision for that, mv + m(0) = mv' + mv'. mv = m(2v'). So velocity after impact v' = v/2. |

Ambrishnitp said: (Mar 3, 2014) | |

One of the body which was moving initially with velocity V would come to rest and second body will move with velocity v. mv +m *0 = m*0 + mv. For perfectly elastic collision velocity gets exchanged and e = 1. |

Kedar Gupta said: (Apr 22, 2014) | |

After collision the second body will move by velocity 'v' because first body will remain rest and second will move. |

Raja Tms said: (May 18, 2014) | |

We know that total momentum remains constant i.e., m1*v1=m2*v2 given that mass is constant m1=m2 then v1=v2 so the 1st ball gives half of its velocity to 2nd. |

Sampad Kumar Rout said: (Jun 6, 2014) | |

For perfectly elastic body coefficient of restitution =1 so first body will stop and 2nd will move in a velocity v. |

Satish said: (Jun 25, 2014) | |

From conservation of momentum. Momentum before impact = momentum after impact. mv+m*0 = mv1+mv1 { after impact v1=v2}. Hence, v1 = v/2. |

Muluh Theodore said: (Jun 29, 2014) | |

Let the initial velocity of first mass be u. Final velocities attained by first and second masses be v and w respectively. From conservation of momentum, mu = mv + mw We can have v = u - w ........(1). From conservation of kinetic energy, 1/2mu2 = 1/2mv2 + 1/2mw2 NB: u2 is read u square. u2 = v2 + w2................(2). (1) in (2), u2 = (u - w)2 + w2. simplifying gives 2uw = 2w2. w = u. in (1), v = 0. This simply means that when the first mass collides with the second, it comes to rest and the second mass takes off with the initial velocity of the first. The analysis is simple. This was proven by Newton and the experiment he used is called Newton's Craddle. |

Sadam Hussain said: (Jul 5, 2014) | |

Initial momentum = final momentum. |

Nazik said: (Aug 24, 2014) | |

mv1_+mv2 = mv'1+mv'2. m is equal. v1+0 = 2v' because v1 = 0 and v'1 = v'2. v' = v/2. |

Satyajit Sahu said: (Oct 14, 2014) | |

As here one velocity is responsible for motion of two equal mass body so, the velocity will distribute. As we know the co-efficient of restitution(e=1) for perfectly body. mv = mv1+mv2. mv = mv'+mv'. v' = v/2. |

Santhosh.M.R said: (Oct 29, 2014) | |

Two balls have same mass and elastic property. So e = 1, Momentum before impact = momentum after impact. mv+m*0 = mv1+mv1 {after impact v1 = v2}. Hence, v1 = v/2. |

Shivam said: (Nov 26, 2014) | |

It should be perfectly inelastic material then only V/2 answer will be correct. |

Parth said: (Nov 27, 2014) | |

Its wrong, only V2 has velocity of V after impact. |

Dinesh said: (Apr 16, 2015) | |

After striking the velocity of the two bodies will be reduces why because assume both the two bodies are rotate in CW direction when second body front side touches the first body back side due to same direction speed will be reduces. |

Raviteja Reddy said: (Apr 23, 2015) | |

Law of conservation of momentum. |

Sumeet said: (May 5, 2015) | |

Law of conservation of momentum. When one ball hits second then the first ill stop & its momentum is transfer to other with the half of the velocity of first ball. That's why answer = V/2. |

Sanket Haste said: (Jun 14, 2015) | |

Initial velocity = v. Given condition that both balls are moving. Hence v/2. |

Jay said: (Jun 16, 2015) | |

For two perfectly elastic body, the velocity after collided become same i.e. velocity 1st ball after collide equal to velocity of 2nd ball after collide. So final velocity of the ball half of the initial velocity. It will get from the equation of conservation of momentum after and before. |

Vivek Kumar Singh Ideal Ghaziabad said: (Jul 30, 2015) | |

According to question balls is perfectly elastic material then the total energy of first ball is transfer to second ball after collision then first ball energy less it means no energy no movement then the velocity of first ball is 0 and second ball is travel of velocity v. |

Deepak said: (Jan 5, 2016) | |

The total energy of momentum of force is directly impact v = 1/v, it is impact force. |

Uzzal said: (Mar 30, 2016) | |

mv1_+mv2 = mv'1+mv'2. m is equal. v1+0 = 2v' because v1 = 0 and v'1 = v'2. v' = v/2. |

Madheswaran Bit said: (Apr 15, 2016) | |

Velocity after striking will be v/2. Because it is a perfectly elastic body so after striking both are will move in an opp direction so by applying momentum equation. Total momentum before striking = Total momentum After strike. m(v1i+v2i) = m(v1f+v2f). @ first V2i = 0. mV = m(v1+V2) Perfectly elastic so V1f=V2F. V=2Vf. V(final)=V/2. |

Bantee Raja Bundela said: (May 6, 2016) | |

mv1 + mv2 = mv1' + mv2' Where m = mass (same for both the balls in this case) v1 = v (velocity of moving ball, given) v2 = 0 (velocity of stationary ball, given) => mv + m(0) = mv' + mv'. mv = m(2v'). So, v1' = v2' = v/2. |

Ravi said: (May 10, 2016) | |

According to the conservation of momentum. The 1st ball will come to rest and the second ball will move with a velocity V. |

Pavan said: (Jul 22, 2016) | |

mv1 + mv2 = mv'1+mv'2. m is equal. v1 + 0 = 2v' because v1 = 0 and v'1 = v'2. v' = v/2. |

Sagar Gupta said: (Jul 27, 2016) | |

For perfectly elastic body the velocity of each body is v/2. |

Sohan said: (Aug 31, 2016) | |

After impact, the ball 1 will come into rest and 2 will move with velocity we. What you guys are saying that happens only if the collision is perfectly plastic? |

Vikas said: (Sep 3, 2016) | |

Let the initial velocity of first mass be u. Final velocities attained by first and second masses be v and w respectively. From conservation of momentum, mu = mv + mw, We can have w = u - v ----->(1). From conservation of kinetic energy, 1/2 mu2 = 1/2 mv2 + 1/2 mw2, u2 = v2 + w2 ----->(2). (1) in (2), u2 = v2 + (u - v)2. Simplifying gives 2 uv = 2 v2. v = u. In (1), w = 2u. After collision ball 1 return with the same velocity and other attain a twice velocity. |

Vikas said: (Sep 3, 2016) | |

Let the initial velocity of first mass be u. Final velocities attained by first and second masses be v and w respectively. From conservation of momentum, mu = mv + mw, We can have w = u - v ----->(1). From conservation of kinetic energy, 1/2 mu2 = 1/2 mv2 + 1/2 mw2, u2 = v2 + w2 ----->(2). (1) in (2), u2 = v2 + (u - v)2. Simplifying gives 2 uv = 2 v2. v = u. In (1), w = 2u. After collision ball 1 return with the same velocity and other attain a twice velocity. |

Mahantesh Jakkappanavar said: (Sep 24, 2016) | |

It is newtons second law. |

Pritam Samanta said: (Feb 21, 2017) | |

We know co-efficient of restitution, e=(u1-v1)/(v2-u2) Where u1,u2 are the initial velocities and v1,v2 are the final velocities. For perfectly elastic material e=1, And u1=0, u2= v and v1=v2, So the eq becomes, 1=(0-v1)/(v2-v). or, v2-v=-v1. or, 2v1=v. or, v1=v2=v/2. |

Dhanendra Sahu said: (Jun 24, 2017) | |

Velocity of the first particle of same mass after collision gets 0 and vel of sec particle after collision assumes the velocity of the first particle before collision( ie. V2 =U1, V1=0, U2=0,) For a perfect elastic body property. |

Johirul said: (Aug 6, 2017) | |

Ans. : V/2 is correct. |

Suraj said: (Sep 28, 2017) | |

@ALL. It is the fact is it is a perfectly elastic collision so no kinetic energy losses. So the velocity of the 1st body will transfer to next body. And the answer will be V. |

Dhruv said: (Jan 2, 2018) | |

We can not take velocity only based on momentum. we need to take in condition loss of kinetic energy which is zero in this case. So we have two equation V = V1 + V2 --------------> 1 Square of V =Square of V1 +Square of V2 --------------> 2 Solve this, you will get V1=0 and V2=V. |

Sk Yadav said: (Jan 5, 2018) | |

If the second mass it at rest initially. After collision 1st mass gets rest second mass move with a velocity ' V'. NOT By both ' V/2'. |

Sandeep Chaudhary said: (Jan 7, 2018) | |

Cconservation of momentam. mv=m1v1+m2v2. mv=mv+mv. v=v/2. |

Sakthivel said: (Jan 9, 2018) | |

Conservation of momentum Consider U-initial velocity V-final velocity m1u1+m2u2=m1v1+m2v2 m(u1+0)=m1(v1+v2) u1=v1+v2 Therefore finial velocity equally distributed So u=2v. V=u/2. |

Aman said: (Feb 17, 2018) | |

See, there is a concept of coefficient of restitution(e). i.e e= Velocity of separation/ velocity of approach. While here velocity of approach is V. and for the perfectly elastic case, e=1; Now, let's assume after impact velocity of ball 1 and ball 2 id V1 and V2. Then, by the coefficient of restitution concept: velocity of separation = velocity of approach. i.e V = V2 - V1 -----> (1) and by conservation of momentum; mV = mV1 + mV2 => V1 + V2 = V -----> (2) after solving eq(1) and eq (2); you will get V2 = V and V1 =0 ; In Nutshell, after collision ball 1 will come to rest and ball 2 will gain the velocity 'V'. |

Rahul said: (Feb 21, 2018) | |

Aman is right but in question given that after collision both balls move with the same velocity then use that concept in momentum eq; mv = mv1+mv1, so v1=v/2. |

Ravan said: (May 17, 2018) | |

@ALL. As per my knowledge, after the collision, both balls are moving with some velocity so that the velocity of balls after the collision is v/2. |

Mithun Kumar Biswas said: (Jun 9, 2018) | |

I think it would be v/√2. If the second body before struck were in rest. Must answer should be v/√2. (kinetic energy should be equal before and after collision). |

Wravin111 said: (Jul 5, 2018) | |

m1u1+m2u2=(m1+m2)v2. u2=0 as it's stationary. So, mu/2m=v2, 1u/2=v2. |

Ajeesh said: (Nov 21, 2018) | |

This is considered as inelastic collision so both will move in same velocity, V/2, If the collision was in the vacuum, ie perfectly elastic collision, V1F will be 0 and V2F will be V. |

Anoima said: (Mar 1, 2019) | |

I think only the second mass will move with velocity we and the first mass will get into rest. |

Dhruvil Prajapati said: (Jul 21, 2019) | |

The answer given is incorrect. Reason: A perfectly elastic ball on striking with another perfectly elastic ball will cause the second ball to move with velocity with which the first ball was approaching and the first ball will come to rest after the collision. Both Energy, as well as momentum conservation, has to be applied. Total Kinetic Energy (T.E) = 1/2 mv1^2 Final K.E = T.E Therefore, 1/2mv1^2 = 1/2mv1'^2 + 1/2mv2'^2 ------> (1) Also mv1 = mv1' + mv2' ------>(2) From (2) v1 = v1' + v2' ------> (3) Substituting (3) in (1) (v1' + v2')^2 = v1'^2 + v2'^2. v1'^2 + 2v1'v2' + v2'^2 = v1'^2 + v2'^2. v1'v2' = 0. So, either v1'= 0 or v2' = 0. But since v2' cannot be zero after the collision (since m2 receives an impulse from m1) Hence, v1' = 0. On substituting the value of v1' in eqn (3) v2' = v1. Hence Proved. |

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