Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 3)
3.
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
Discussion:
70 comments Page 2 of 7.
Dhruv said:
8 years ago
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.
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.
Vivek kumar singh ideal ghaziabad said:
1 decade ago
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.
Pritam Samanta said:
9 years ago
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.
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.
Dinesh said:
1 decade ago
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.
Dhanendra sahu said:
8 years ago
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.
For a perfect elastic body property.
BANTEE RAJA BUNDELA said:
9 years ago
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.
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.
Satyajit Sahu said:
1 decade ago
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.
As we know the co-efficient of restitution(e=1) for perfectly body.
mv = mv1+mv2.
mv = mv'+mv'.
v' = v/2.
Ambrishnitp said:
1 decade ago
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.
mv +m *0 = m*0 + mv.
For perfectly elastic collision velocity gets exchanged and e = 1.
Shrikant said:
1 decade ago
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.
Abhishek said:
1 decade ago
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.
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