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A 0.001kg bullet is fired with a velocity of 800m/s into a soft wood of mass 1kg resting on a smooth surface. Find the final vel

V125BC [204]

The final velocity of the bullet+block is 0.799 m/s

Explanation:

We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.

Mathematically, we can write:

$mu+MU=(m+M)v$

where

m = 0.001 kg is the mass of the bullet

u = 800 m/s is the initial velocity of the bullet

M = 1 kg is the mass of the block

U = 0 is the initial velocity of the block (initially at rest)

v is the final combined velocity of the bullet and the block

Solving the equation for v, we find the final velocity:

$v=\frac{mu}{m+M}=\frac{(0.001)(800)}{0.001+1}=0.799 m/s$

Learn more about conservation of momentum:

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#LearnwithBrainly

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3 years ago

Which of the following is an example of an endothermic reaction?

il63 [147K]

Answer:

D, photosynthesis

Explanation:

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As players speed up, their kinetic energy will?

igomit [66]

Kinetic energy is proportional to the square of the speed. So when anything or anybody speeds up, its kinetic energy increases.

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3 years ago

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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

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3 years ago

The launch speed of a projectile is three times the speed it has at its maximum height.what is the elevation angle at launch?

Sphinxa [80]

Answer:

Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.

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2 years ago