Sunday, August 12, 2012

CBSE IX Term -1 Solved numerical Gravitation and Laws of motion

IX Term -1 Solved numerical Gravitation and Laws of motion
Q.A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.003 s to move through its barrel, calculate the force experienced by the rifle due to the recoil.
Ans:    Mass of the rifle (m1) = 3 kg    Mass of the bullet (m2) = 0.03 kg
Initial velocity of the rifle (u1) = u2 = 0,   Final velocity of the rifle (v1) = ?
Final velocity of the bullet (v2) = 100 m s-1
According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
  
Using the law of conservation of momentum,  0 + 0 = 3v1 + 0.03 x 100 =-1m/s 
Negative sign shows that the rifle moves in a direction opposite to that of the bullet,
Force experienced by the rifle due to its recoil= m1v1/t =(3x-1)/0.003=-1000N 
Therefore, the person would experience a force of 1000 N in the backward direction due to the recoil of the rifle.
 Q. Action and reaction forces is never equal to zero. Why?
Ans: The action and reaction force are equal and opposite but their resultant is never equal to zero or cancel because The action and reaction force are acting on two different bodies.
Q. A bullet leaves a rifle with a velocity of 100m/s and the rifle of mass 2.5 kg recoils with a velocity of 1m/s. Find the mass of the bullet?
Ans: According to law of conservation of momentum = m1u1 + m2u2 = m1v1 + m2v2
Þ 0=2.5x1 +m2 x100  Þ 2.5/100= the mass of the bullet  Þ 25gm= the mass of the bullet
Q. A body of mass 5 kg undergoes a change in speed from 30 to 40m/s . Calculate its increase in momentum?
Ans: Increase in momentum =m(v-u) =5 kg(40-30)m/s=50kgm/s
Q. Two bodies of mass 10 kg and 12 kg are falling freely. What is the acceleration produced in the bodies due to force of gravity?
Ans: 9.8 m/s2.ast he acceleration due to gravity produced in both the bodies is the same as it is independent of the mass of the body.
Q. What will happen to the force of gravitation between two objects A and B if the distance between them is reduced to half?
Ans: F=1/r2
F’ = 1/(1/2r)2 =4(1/r2) =1/4(F)
if the distance between two objects A and B is reduced to half the force of gravitation between two objects A and B increases 4 times
Q. What will happen to the force of gravitation between two objects A and B If the mass of the object A is doubled.
Ans: F = m M
F’= 2mxM=2(mM)=2F
the force of gravitation between two objects A and B increases 2 times
Q. Show that universal gravitational constant is nothing but force of gravitation between two unit masses separated by unit distance.
Ans: F=GmM/r2
if m=M=1kg and r =1m
then, F = G i.e., gravitational constant is equal to the force of gravitation.
Q. A boy drops a stone from a cliff, which reaches the ground in 20 seconds. Calculate the height of the cliff.
Ans: the height of the cliff.=S=ut+1/2at2 =0+1/2x10x20x20=2000m=2km

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