Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side smooth surface…..

Q: Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side smooth surface. A bullet of mass m = 20 gm strikes the block M1 and pierces through it, then strikes the second block and sticks to it. Consequently both the blocks move with equal velocities. Find the percentage change in speed of the bullet when it escapes from the first block.

Sol: Conservation of linear momentum of the system yields

$\displaystyle m v_0 = M_1 v + m v’ = (M_2 + m ) v $

Numerical

$\displaystyle v = \frac{m v_0}{M_2 + m}$

and mv’ = m v0 – M1 v

$\displaystyle m v’ = m v_0 – \frac{M_1 m v_0}{M_2 + m}$

$\displaystyle m v’ = m v_0(1 – \frac{M_1}{M_2 + m}) $

Now find , $\displaystyle \frac{|v’-v_0|}{v_0} \times 100 = \frac{M_2 – M_1 + m}{M_2 + m} \times 100 $

A cannon and a supply of cannon balls are inside a sealed railroad car. The cannon fires to the right, …

Q: A cannon and a supply of cannon balls are inside a sealed railroad car. The cannon fires to the right, the car recoils to the left. The cannon balls remain in the car after hitting the far wall. Show that no matter how the cannon balls are fired, the railroad car cannot travel more than l , assuming it starts from rest .

Sol: The speed of the car = vc The speed of the balls = vb

m & M are the mass of each ball and the rest of the given system respectively.

Numerical

$\displaystyle \vec{m v_b} + \vec{M v_c} = 0 $ ; Since $F_{ext} = 0 $

$\displaystyle m (\vec{v_{bc}} + \vec{v_c}) + \vec{M v_c} = 0 $

$\displaystyle \vec{v_c} = – \frac{\vec{m v_{bc}}}{M+m} $ ; ; vbc =velocity of ball relative to car.

$\displaystyle x_c = \int_{0}^{t}v_c dt $

$\displaystyle x_c = \frac{m}{M+m} \int_{0}^{t}v_{bc} dt $

$\displaystyle x_c = \frac{m l}{M+m}$

⇒ xc < l

Where xc distance travelled by the car and

$\displaystyle \int_{0}^{t}v_{bc} dt $ = total distance travelled by each ball relative to the car = l

Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane…

Q: Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane. There exists friction between A and B. If a bullet of mass m hits the lower block with a horizontal velocity v and gets embedded into it. Find the work done by friction between A and B.

Sol: Let the velocity of lower block just after collision be v1,

Numerical

Applying COLM between bullet and lower block as impulse due to friction is negligible.

$\displaystyle m v = (M + m)v_1 $

$\displaystyle v_1 = \frac{m v}{M+m} $

After collision frictional force acts on both and will act till velocities of both blocks become equal.

$\displaystyle (M+m)v_1 = (2M+m)v_2$

$\displaystyle v_2 = \frac{(M+m)v_1}{2M+m} $

Applying work energy theorem ,

$\displaystyle W_{friction }= K_f -K_i$

$\displaystyle W_{friction }= \frac{1}{2}(2M+m)v_2^2 – \frac{1}{2}(M+m)v_1^2$

$\displaystyle W_{friction }= -\frac{M m^2 v^2}{2(m+2M)(m+M)}$

A block of mass M is hanging from a rigid support by an in-extensible light string…..

Q: A block of mass M is hanging from a rigid support by an in-extensible light string. A ball of mass m hits it with a vertical velocity v at its bottom. Find the change in momentum of the ball assuming inelastic collision.

Sol: Let the velocity of combined mass of bullet and block just after collision be v1
Applying COLM,
$\displaystyle m v = (M + m) v_1 $

$\displaystyle v_1 = \frac{m v}{M+m}$
Numerical

Change in the momentum of bullet = Pf – Pi = mv1 – mv

$\displaystyle = \frac{m^2 v}{M+m} – m v $

$\displaystyle = – \frac{M m }{M+m}v $

A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface….

Q: A block of mass M with a semicircular track of radius R rest on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A. The cylinder slips in the semicircular frictionless track. How far the block moved when the cylinder reaches the bottom of the track?

Sol: Since no net horizontal force acts on the system, it conserves its horizontal momentum.

$\displaystyle \vec{P} = M \vec{v_b} + m \vec{v_c} = 0 $ .

Because initially the system was stationary.

Numerical

$\displaystyle M \vec{v_b} + m \vec{v_c} = 0 $ , Where $\vec{v_b}$ & $\vec{v_c}$ are the horizontal component of velocities of the block and the cylinder respectively

vcb = velocity of the cylinder relative to the block = v (say)

$\displaystyle v_b = \frac{m v_{cb} sin\theta}{M + m}$ ( Numerically )

$\displaystyle \int_{0}^{t} v_b dt = \frac{m}{M+m}\int_{0}^{t} (v_{cb})_x dt $

where t = time taken by the cylinder to arrive at the bottom of the block& (vcb)x = horizontal component of velocity of the cylinder relative to the block.

$\displaystyle x_b = \frac{m x_{cb}}{M+m} $ ; putting xbc = (R-r) we obtain

$\displaystyle x_b = \frac{m}{M+m}(R-r) $ ; Where $\displaystyle \int_{0}^{t} (v_{cb})_x dt = (R-r)$ and $\displaystyle \int_{0}^{t} v_b dt = x_b $