Physics, 07.11.2019 04:31, jfy3341

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. the catapults are placed very close to a pendulum bob of mass 10.0 kg attached to a massless rod of length l. assume that catapult 1 fires a projectile of mass 5.50 g at speed v1 and catapult 2 fires a projectile of mass 10.0 g at speed v2. if the projectile from catapult 1 causes the pendulum to swing to a maximum angular displacement of 6.80 ∘ and the projectile from catapult 2 causes a displacement of 11.4 ∘, find the ratio of the initial speeds, v1v2 .

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The conservation of momentum and energy allows finding the result for the relationship of the velocities of the balls fired by the two catapults is:

$\frac{v_1}{v_2} = 1.0 85$

Given parameters

The mass of the pendulum M = 10 kg. The mass of each ball m₁ = 5.50 g = 5.50 10⁻³ kg and m₂ = 10.0 g. pendulum angle te1 = 6.80º and tea2 = 11.4º

To find

The ratio of ball speeds v1 / v2

The momentum is defined by the product of the massby the speed of the particle, without the system is isolated the momentum is conserved.

P = mv

Where bold letters indicate vectors, p is momentum, m is mass, and v is velocity.

Let's solve the problem for each ball separately, let's start with case 1 the ball of m₁ = 5.50 10⁻³ kg.

The system is shaped by the ball and the pendulum, therefore it is an isolated system and the momentum is conserved.

Initial instant. Before crash

p₀ = m₁ v₁

Final moment. Right after the crash

p_f = (m₁ + M) v

The momentum is preserved

p₀ = p_f

m₁ v₁ = (m₁ + M) v

v₁ = $\frac{m_1 + M}{m_1} \ v$ (1)

Now we can use the conservation of energy for the pendulum with the embedded ball m1, see attached.

Starting point. Just when they're coming out at the bottom

Em₀ = K = ½ (m₁ + M) v²

Final point. The highest part of the movement

Em_f = U = (m₁ + M) gh

Energy is conserved.

Em₀ = Em_f

½ (m₁ + M) v² = (m₁ + M) gh

v = $\sqrt{2gh}$

Let's use trigonometry to find the height of the pendulum, see attached

h = L - L cos θ₁

Let's substitute

v = $\sqrt{2gL ( 1 - cos \theta_1)}$

Let's substitute in equation 1

$v_1 = \frac{m_1+M}{m_1} \ \sqrt{2g L ( 1- cos \theta_1) }$

We solve the case 2 ball of mass m₂ = 10 10⁻³ kg.

The system has the same approach: the conservation of the moment and then the conservation of energy, therefore the expression for the speed of this ball is:

$v_2 = \frac{m_2+M}{m_2} \ \sqrt{2gL(1-cos \theta_2) }$

They ask us for the relationship between the two speeds.

Let's calculate each velocity and then make the relationship.

Suppose a pendulum length value of L = 1 m, to make an explicit calculation.

$v_1 = \frac{0.0055 + 10}{0.0055} \ \sqrt{2 \ 9.8 \ 1 (1 - cos 6.8)}$

v₁ = 675.49 m / s

$v_2 = \frac{0.01 + 10}{0.010} \ \sqrt{2 \ 9.8 \ 1(1 - cos 11.4)}$

v₂ = 622.46 m/s

The ratio of the speeds is:

$\frac{v_1}{v_2} = \frac{675.49}{622.46}$

$\frac{v_1}{v_2}$ = 1,085

In conclusion, using the conservation of momentum and energy, we can find the relationship of the velocities of the balls fired by the two catapults is:

$\frac{v_1}{v_2}$ = 1,085

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An experiment is done to compare the initial speed of projectiles fired from high-performance catapu

answered: phil820

1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

answered: Guest

1 the 4 is sig

answered: Guest

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

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An experiment is done to compare the initial speed of projectiles fired from high-performance catapu...