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Answer:
f1/f2 =W1/W2 = 1/3
.0 f2 = 3f1
As ,
1/F= 1/f1 +1/f2
...1/40 = 1/f1 - 1/3f1
f1=> 80/3 cm
... f2 = 2f1 = 3 x 80/3 = 80 cm
Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
In Newton's Cradle experiment we know that all cradles of same mass and identical to each other
Now we know that when two identical objects collide elastically then they interchange their velocity
So here we have same illustration
When Newton pulls up a cradle and release it will move hit another cradle which is at rest
Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.
So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.
So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.
So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.
