
Two wooden planks of mass 1 kg and 2.98 kg having a smooth surface. A bullet of mass 20 g strikes the first block and pierces through it, then strikes the plank B and sticks to it. Consequently, both the plank moves with equal velocities. If the percentage change in speed of the bullet when it escapes from the first block is x/3 %, find the value of x.
Answer
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Hint: Since, there is no external force acting the total momentum of the system must be conserved that is total momentum before Collision must be equal to the total momentum after collision. Momentum depends upon the variables mass and velocity of the body involved.
Complete step by step answer:
Initially only the bullet was moving and the two planks A & B were at rest. After the collision the bullet sticks to plank B and both the planks move with the same velocity, v. There are two collisions, so we can use conservation of momentum twice.
Mass of bullet= 20g = 0.02 kg
Mass of plank A= 1 kg
Mass of plank B = 2.98 kg
Let the initial velocity of the bullet be u, after it emerges from the plank A let it be u’ and let the velocities of the two planks be v.
Using the law of conservation of the momentum for collision between bullet and plank A: $mu={{M}_{A}}v+mu'$---(1)
Using the law of conservation of the momentum for collision between bullet and plank B: $mu=({{M}_{B}}+m)v$---(2)
From eq (2), $v=\dfrac{mu}{({{M}_{B}}+m)}$---(3)
From eq (1), $mu'=mu-{{M}_{A}}v$, substituting the value of v from (3) we get,
$mu'=mu-\dfrac{{{M}_{A}}mu}{({{M}_{B}}+m)}$
$\implies mu'=mu\{1-\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\}$
$\implies u'=u\{1-\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\}$
$\implies u'=u-\dfrac{{{M}_{A}}u}{({{M}_{B}}+m)}$
$\implies \left| \dfrac{u'-u}{u} \right|\times 100=\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\times 100=\dfrac{1}{3}\times 100=33.33\%$
But it was given to be $\dfrac{x}{3}\%$
$ \dfrac{x}{3}\%$ = $ \dfrac{100}{3}\%$
$\therefore X= 100$
So, the value of X comes out to be 100.
Note:
This is the easiest way to solve this problem. Momentum conservation is a universal law which holds until there are no external force acts on the system. Also, there is no loss of energy taking place in the system. All the units must be in standard SI convention.
Complete step by step answer:
Initially only the bullet was moving and the two planks A & B were at rest. After the collision the bullet sticks to plank B and both the planks move with the same velocity, v. There are two collisions, so we can use conservation of momentum twice.
Mass of bullet= 20g = 0.02 kg
Mass of plank A= 1 kg
Mass of plank B = 2.98 kg
Let the initial velocity of the bullet be u, after it emerges from the plank A let it be u’ and let the velocities of the two planks be v.
Using the law of conservation of the momentum for collision between bullet and plank A: $mu={{M}_{A}}v+mu'$---(1)
Using the law of conservation of the momentum for collision between bullet and plank B: $mu=({{M}_{B}}+m)v$---(2)
From eq (2), $v=\dfrac{mu}{({{M}_{B}}+m)}$---(3)
From eq (1), $mu'=mu-{{M}_{A}}v$, substituting the value of v from (3) we get,
$mu'=mu-\dfrac{{{M}_{A}}mu}{({{M}_{B}}+m)}$
$\implies mu'=mu\{1-\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\}$
$\implies u'=u\{1-\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\}$
$\implies u'=u-\dfrac{{{M}_{A}}u}{({{M}_{B}}+m)}$
$\implies \left| \dfrac{u'-u}{u} \right|\times 100=\dfrac{{{M}_{A}}}{({{M}_{B}}+m)}\times 100=\dfrac{1}{3}\times 100=33.33\%$
But it was given to be $\dfrac{x}{3}\%$
$ \dfrac{x}{3}\%$ = $ \dfrac{100}{3}\%$
$\therefore X= 100$
So, the value of X comes out to be 100.
Note:
This is the easiest way to solve this problem. Momentum conservation is a universal law which holds until there are no external force acts on the system. Also, there is no loss of energy taking place in the system. All the units must be in standard SI convention.
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