Laws of Motion
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Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration ‘a’, then the acceleration of mass B will be:
Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration ‘a’, then the acceleration of mass B will be:
Let the spring is in extension state and
Hence we can say that block moves away from block B in the frame of B
F – kx = MaB . (1)
kx = Ma &nb
Upvote 
Kindly consider the following figure
Upvote A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at
t = 6 s ? (Take g = 10 m/s² )
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A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at
t = 6 s ? (Take g = 10 m/s² )
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velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s
Upvote A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:
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A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:
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By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF
Upvote A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of θ is approximately: (take g = 10 m/s²)
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A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of θ is approximately: (take g = 10 m/s²)
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By Conservation of Angular Momentum (COAM) about O, just before and after the collision:
Initial angular momentum L? = mv?
Final angular momentum L? = Iω? = (I_rod + I_block)ω? = (M? ²/3 + m? ²)ω?
L? = L?
mv? = (M? ²/3 + m? ²)ω?
ω? = mv / (M? /3 + m? ) = (16) / (21/3 + 1*1) = 6 / (5/3) = 18/5 rad/s
By
Upvote A particle of mass m with an initial velocity ui collides perfectly elastically with a mass 3m at rest. It moves with a velocity vj after collision, then v is given by
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A particle of mass m with an initial velocity ui collides perfectly elastically with a mass 3m at rest. It moves with a velocity vj after collision, then v is given by
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From momentum conservation
mui + 0 = mvj + 3mv'
v' = u/3 I - v/3 j
From kinetic energy conservation 1/2 mu² = 1/2 mv² + 1/2 (3m) (u/3)²+ (v/3)²)
Solving, v = u/√2
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