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.
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.
Mahantesh Jakkappanavar said:
9 years ago
It is newtons second law.
Vikas said:
9 years ago
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.
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:
9 years ago
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.
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.
Sohan said:
9 years ago
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?
What you guys are saying that happens only if the collision is perfectly plastic?
Sagar gupta said:
9 years ago
For perfectly elastic body the velocity of each body is v/2.
Pavan said:
9 years ago
mv1 + mv2 = mv'1+mv'2.
m is equal.
v1 + 0 = 2v' because v1 = 0 and v'1 = v'2.
v' = v/2.
m is equal.
v1 + 0 = 2v' because v1 = 0 and v'1 = v'2.
v' = v/2.
Ravi said:
9 years ago
According to the conservation of momentum.
The 1st ball will come to rest and the second ball will move with a velocity V.
The 1st ball will come to rest and the second ball will move with a velocity V.
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.
Madheswaran BIT said:
9 years ago
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.
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.
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