2) Two pitch ball apart by 3.2×10^2M
are released from rest, the alt
acceleration of the first ball
is 7. Qm 15 and the second ball
is 9.0 m/s2, If mass m1= 6.3x10-7 kg
What are the (a) the mass of
the second mass m2? (b) the
magnitude of charge of each
particle?
Answers
Answer:
According to Work energy theorem
for light mass ball,velocity v
1
ΔE
K
=W where ΔE
K
=Kinetic energy and W=work done.
2
1
mv
1
2
−
2
1
m0
2
=mgl
v
1
=
2gl
Similarly for heavy ball,velocity v
2
=
2gl
Hence v
1
=v
2
=x----------(A)
As per law of conservation of momentum,
mv
1
−2mv
2
=mv
1
+2mv
2
mx−2mx=mv
1
+2mv
2
−x=v
1
+2v
2
-----------------(B)
Again
v
1
−v
2
=−(v
1
−v
2
)
v
1
−v
2
=−2x as v
1
=v
2
=x---------------(C)
Subtracting B and C, we get
3v
2
=x
v
2
=
3
x
=
3
2gl
---------------(D)
Substituting the above value in c we get,
v
1
=−2x+
3
x
=
3
−5x
=
3
−5
2gl
----------(E)
(b). Again from work energy theorem we get,
2
1
2m0
2
−
2
1
2mv
2
2
=2mgh
Putting the value of D in the above equation,
9
gl
=gh
h=
9
l
The above h=height for the heavy ball
Similarly for light ball,
2
1
m0
2
−
2
1
mv
1
2
=mgh
Putting the value of E in the above equation,
h
1
=
9
25
l
Answer:
According to Work energy theorem
for light mass ball,velocity v
1
ΔE
K
=W where ΔE
K
=Kinetic energy and W=work done.
2
1
mv
1
2
−
2
1
m0
2
=mgl
v
1
=
2gl
Similarly for heavy ball,velocity v
2
=
2gl
Hence v
1
=v
2
=x----------(A)
As per law of conservation of momentum,
mv
1
−2mv
2
=mv
1
+2mv
2
mx−2mx=mv
1
+2mv
2
−x=v
1
+2v
2
-----------------(B)
Again
v
1
−v
2
=−(v
1
−v
2
)
v
1
−v
2
=−2x as v
1
=v
2
=x---------------(C)
Subtracting B and C, we get
3v
2
=x
v
2
=
3
x
=
3
2gl
---------------(D)
Substituting the above value in c we get,
v
1
=−2x+
3
x
=
3
−5x
=
3
−5
2gl
----------(E)
(b). Again from work energy theorem we get,
2
1
2m0
2
−
2
1
2mv
2
2
=2mgh
Putting the value of D in the above equation,
9
gl
=gh
h=
9
l
The above h=height for the heavy ball
Similarly for light ball,
2
1
m0
2
−
2
1
mv
1
2
=mgh
Putting the value of E in the above equation,
h
1
=
9
25
l
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