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otez555 [7]
4 years ago
8

The two blocks of masses M and 2M shown above initially travel at the same speed v but in opposite directions. They collide and

stick together. What is the final velocity of the mass M after the collision?a. -1/6v.b. -1/3v.c. 1/6v.d. 1/3v.Figure:two box are colliding
Physics
1 answer:
Vinil7 [7]4 years ago
6 0

Answer:

option B.

Explanation:

given,

Mass of the blocks 1 = M

Mass of the blocks 1 = 2 M.

initial speed of the block 1= v

initial speed of the block 2 = - v

where V is the final speed of the blocks after collision

using conservation of momentum

M v + 2 M (-v) =  (M + 2 M) V

3 M V = - M v    

V = -\dfrac{v}{3}      

velocity of the blocks after collision is equal to -v/3 m/s

hence, the correct answer is option B.

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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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Rudik [331]
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What is the wavelength of a photon whose energy is twice that of a photon with a 600 nm wavelength?
bearhunter [10]
Planck's equation states that
E = hf
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Because
c = fλ
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c =  velocity of light,
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E = h(c/λ)

Photon #1:
The wavelength is λ₁ = 60 nm.
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E₁ = (hc)/λ₁

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E₂ = 2E₁ = (hc)/λ₂.

Therefore
\frac{E_{2}}{E_{1}}= \frac{(hc)/\lambda_{2}}{(hc)/60 \, nm} =2\\ \frac{60}{\lambda_{2}} =2 \\ \lambda_{2} =  \frac{60}{2} =30 \, nm

Answer:  30 nm

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Answer:

Explanation:

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