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A particle of mass m moving with velocity u_{1} collides elastically with another particle of same mass moving with velocity u_{2} in the same direction. After collision their speeds are v_{1} and v_{2} respectively then
(A) u_{1} + v_{1} = v_{2} + u_{2}
(B) u_{1} – v_{1} = v_{2} + u_{2}
Equation A is correct but not B
(For elastic collision e = 1)
⇒ 1(u_{2} – u_{1}) = –(v_{2} – v_{1})
⇒ u_{2} + v_{2} = u_{1} + v_{1}
The correct answer is: Equation A is correct but not B
The centre of the mass of the shaded portion of the disc is : (The mass is uniformly distributed in the shaded portion) :
The correct answer is: R/20 to the left of A
A man of mass M hanging with a light rope which is connected with a balloon of mass m. The system is at rest in air. When man rises a distance h with respect to balloon. Find the distance raised by man :
Initially
since no external force is acting
∴ COM should be at rest
Let balloon descend by a distance x.
Mh = (m + M)x
(Distance descend by balloon)
(Distance raised by man)
The correct answer is:
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is :
According to conservation of energy
MkL^{2} = p^{2} (p = mu)
The correct answer is:
A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The minimum energy released in the process of explosion is :
The correct answer is: (3/2) mv^{2}
Find the position of centre of mass of the uniform planar sheet shown in figure with respect to the origin (O) :
A ball of mass m strikes the fixed inclined plane after falling through a height h. If it rebounds elastically, the impulse on the ball is :
Here e = 1
If ball rebounded elastically
v_{LOI} = u_{LOI}
∴ Along line of impact momentum conservation
Along line of impact
The correct answer is:
The spacecraft of mass M moves with velocity V in free space at first, then it explodes breaking into two pieces. If after explosion a piece of mass m comes to rest, the other piece of space craft will have :
Initial momentum of the system P_{i} = MV
Let the mass of one part be m
Its velocity V = 0
So, mass of other part m′ = M − m
Let its velocity be V′.
Final momentum of the system Pf = mV + m′V′
Or P_{f} = 0+(M−m)V' = (M−m)V′
Applying the conservation of linear momentum P_{f} = P_{i}
∴ (M−m)V' = MV
⟹ V'= MV/(Mm)
A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half the kinetic energy is lost by impact. What is the value of the coefficient of restitution?
The correct answer is: 1/√2
A bullet of mass a and velocity b is fired into a large block of mass c. The final velocity of the system is :
From momentum conservation P_{i} = P_{f}
ab = (a + c)v
The correct answer is:
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