How Do You Know if a Collision Is Elastic or Inelastic

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A 2200kg car travelling at $30\frac{m}{s}$ collides with another 2200kg auto that is at residuum. The two bumpers lock and the cars move forwards together. What is their concluding velocity?

Correct answer:

$15\frac{m}{s}$

Explanation:

This is an example of an inelastic collision, as the 2 cars stick together after colliding. We can assume momentum is conserved.

To make the equation easier, let's call the commencement car "i" and the 2d motorcar "2."

Using conservation of momentum and the equation for momentum, p=mv , nosotros can set upwardly the following equation.

$m_1v_{1initial}+m_2v_{2initial}=(m_1+m_2)v_{final}$

Since the cars stick together, they will have the same concluding velocity. Nosotros know the second car starts at residual, and the velocity of the first auto is given. Plug in these values and solve for the final velocity.

$(2200kg* 30\frac{m}{s})+(2200kg* 0\frac{m}{s})=(2200kg+2200kg)v_{final}$

$66000\frac{kg\cdot m}{s}=(4400kg)v_{final}2$

$\frac{66000\frac{kg\cdot m}{s}}{4400kg}=v_{final}$

$15\frac{m}{s}=v_{final}$

A 10kg ball moving at $13\frac{m}{s}$ strikes a 20kg ball at residuum. Afterward the collision the 10kg brawl is moving with a velocity of $7\frac{m}{s}$ . What is the velocity of the 2d ball?

Right answer:

$3\frac{m}{s}$

Caption:

This is an instance of an elastic standoff. We commencement with two masses and terminate with two masses with no loss of energy.

We tin can use the law of conservation of momentum to equate the initial and final terms.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $\small v_{2final}$ .

$(10kg* 13\frac{m}{s})+(20kg* 0\frac{m}{s})=(10kg*7\frac{m}{s})+(12kg*v_{2final})$

$130\frac{kg\cdot m}{s}+0\frac{kg\cdot m}{s}=(70\frac{kg\cdot m}{s})+(20kg* v_{2final})$

${130\frac{kg\cdot m}{s}}-{70\frac{kg\cdot m}{s}}=20kg\cdot v_{2final}$

${60\frac{kg\cdot m}{s}}=20kg\cdot v_{2final}$

$\frac{60\frac{kg\cdot m}{s}}{20kg}=v_{2final}$

$3\frac{m}{s}=v_{2final}$

A 0.25kg ball strikes a 1.25kg ball at rest. After the collision the 0.25kg ball is moving with a velocity of $18\frac{m}{s}$ and the second ball is moving with a velocity of $9\frac{m}{s}$ . What is the initial velocity of the beginning ball?

Correct answer:

$63\frac{m}{s}$

Explanation:

This is an example of an elastic collision. Nosotros start with two masses and end with ii masses with no loss of energy.

We tin can use the constabulary of conservation of momentum to equate the initial and concluding terms.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $v_{1initial}$ .

$(0.25kg* v_{1initial})+(1.25kg* 0\frac{m}{s})=(0.25kg*18\frac{m}{s})+(1.25kg* 9\frac{m}{s})$

$(0.25kg* v_{1initial})=(4.5\frac{kg\cdot m}{s} )+(11.25\frac{kg\cdot m}{s})$

$v_{1initial}=\frac{15.75\frac{kg\cdot m}{s}}{0.25kg}$

$v_{1initial}=63\frac{m}{s}$

A 12kg ball moving at $37\frac{m}{s}$ strikes a 2d ball at rest. Afterward the collision the 12kg ball is moving with a velocity of $19\frac{m}{s}$ and the 2nd ball is moving with a velocity of $4\frac{m}{s}$ . What is the mass of the second brawl?

Correct answer:

54kg

Explanation:

This is an case of an elastic standoff. We start with two masses and end with two masses with no loss of energy.

We tin utilise the law of conservation of momentum to equate the initial and final terms.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $\small m_{2}$ .

$(12kg* 37\frac{m}{s})+(m_2* 0\frac{m}{s})=(12kg*19\frac{m}{s})+(m_2* 4\frac{m}{s})$

$(444\frac{kg\cdot m}{s})+(0\frac{kg\cdot m}{s})=(228\frac{kg\cdot m}{s})+(m_2 * 4\frac{m}{s})$

$(444\frac{kg\cdot m}{s})-(228\frac{kg\cdot m}{s})=(m_2* 4\frac{m}{s})$

$(216\frac{kg\cdot m}{s})=(m_2* 4\frac{m}{s})$

$\frac{216\frac{kg\cdot m}{s}}{4\frac{m}{s}}=m_2$

54kg=m_2

Which of these would Non be an example of an inelastic collision?

Possible Answers:

2 cars crash into each other, and stop with a loud blindside

A match scrapes a matchbook and bursts into flame

A man is clapping his hands such that they motility with equal, only reverse velocities

Ii hydrogen atoms fuse together to form a helium atom and gamma radiation

Neutrons fuse with hydrogen atoms in a nuclear reactor core such that kinetic energy is conserved

Correct respond:

Neutrons fuse with hydrogen atoms in a nuclear reactor cadre such that kinetic energy is conserved

Explanation:

The divergence between an elastic and an inelastic collision is the loss or conservation of kinetic energy. In an inelastic collision kinetic free energy is not conserved, and volition alter forms into sound, heat, radiations, or some other course. In an rubberband standoff kinetic energy is conserved and does non modify forms.

Remember, total energy and total momentum are conserved regardless of the type of collision; however, while energy cannot be created nor destroyed, it tin change forms.

In the answer options, only one pick preserves the total kinetic free energy. The resulting bang from the car crash, the flame from the lucifer, the audio of hands clapping, and the gamma radiation during hydrogen fusion are all examples of the conversion of kinetic energy to other forms, making each of these an inelastic collision. Only the neutron fusion described maintains the conservation of kinetic free energy, making this an rubberband collision.

Hockey puck A (2kg) travels with a velocity of $40\frac{m}{s}$ to the right when information technology collides with hockey puck B (i.6kg), which was originally at rest. After the collision, puck A is stationary. Assume no external forces are in play and that the momentum of the pucks are conserved. What is the last velocity of puck B afterwards the standoff?

Correct reply:

$50\frac{m}{s}$

Explanation:

Elastic collisions occur when two objects collide and kinetic energy isn't lost. The objects rebound from each other and kinetic energy and momentum are conserved. Inelastic collisions are said to occur when the two objects remain together after the standoff so nosotros are dealing with an elastic collision.

$m_{1}v_{1i}=m_{1}v_{1f}+m_{2}v_{2f}$

Above, the subscripts one and 2 denote puck A and B respectively, and the initial momentum of puck B is nothing, so that term is not included in the equation above.

Plug in initial and final velocities and mass:

$2\left(40\frac{m}{s}\right)=1.6V_{2f}$

$80=1.6V_{2f}$

$\frac{80}{1.6}=V_{2f}$

$V_{2f}=50\frac{m}{s}$

A infinite vehicle, in a round orbit around Earth, collides with a small-scale asteroid that ends up in the vehicle's storage bay. For this standoff

Possible Answers:

But momentum is conserved

Only kinetic energy is conserved

Neither momentum nor kinetic energy is conserved

Both momentum and kinetic free energy are conserved

Correct answer:

Only momentum is conserved

Explanation:

This is an inelastic collision as the two objects stick together and move together with the same velocity. Inelastic collisions conserve momentum, only they do not conserve kinetic free energy.

A bullet with mass 0.25g hits a ballistic pendulum with length and mass 100g and lodges in it. When the bullet hits the pendulum it swings up from the equilibrium position and reaches an bending $30^{\circ}$ at its maximum. Determine the bullet's velocity.

Correct answer:

650m/s

Caption:

Nosotros will demand to start at the end of the situation and piece of work backward in order to determine the velocity of the bullet. At the very end, the pendulum with the bullet reaches its maximum pinnacle and therefore comes to a finish. It has gravitational potential energy. At the bottom of the pendulum correct afterwards the bullet collides with it, it has kinetic energy due to the velocity of the bullet. With the constabulary of conservation of energy, we tin can prepare the kinetic energy of the pendulum correct after the collision equal to the gravitational potential energy of the pendulum at the highest point.

To decide the superlative of the pendulum we volition need to apply trig and triangles to find the height. We know that the pendulum makes a 30-degree angle with the equilibrium position at its maximum tiptop. The length of the pendulum is provided which is the hypotenuse of this triangle. We need to notice the adjacent side of this triangle. We tin can employ cosine to decide this.

$Adjacent=Hypotenusecos\left ( \theta \right )$

$Adjacent=1m\cos \left ( 30 \right )$

Adjacent=0.866m

We tin now subtract this value from the length of the pendulum to determine how high off the ground the pendulum is at its highest point.

1m-0.866m=0.134m

We can now set the kinetic energy of the pendulum correct after the collision equal to the gravitational potential energy of the pendulum at the highest point.

KE=GPE

$\frac{1}2{}mv^{2}=mgh$

The mass is the aforementioned throughout so it falls out of the equation.

$\frac{1}2{}v^{2}=gh$

$v^{2}=2gh$

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8m/s^{2})(0.134m)}$

v=1.62m/s

The pendulum with the bullet was moving 1.62m/s after the collision. We can at present utilize momentum to determine the speed of the bullet before the standoff. Conservation of momentum states that the momentum earlier the collision must equal the momentum after the collision.

$m_{1}v_{i1}+m_{2}v_{i2}=m_{1}v_{f1}+m_{2}v_{f2}$

$m_{1}=0.25g=0.00025kg$

$m_{2}=100g=0.1kg$

They both move together subsequently the standoff

$v_{f1}=v_{f2}=1.62m/s$

Since the pendulum was not moving at the starting time

$v_{i2}=0m/s$

We tin now plug in these values and solve for the missing slice.

$0.00025kgv_{i1}+0=0.00025kg(1.62m/s+0.1kg(1.62m/s)$

$0.00025kgv_{i1}=.162405kgm/s$

$v_{i1}=649.62m/s=650m/s$

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How Do You Know if a Collision Is Elastic or Inelastic

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