Physics, 12.01.2021 18:20 DiiegoWasHere

A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 18.0 cm/s. After the collision, the first marble moves to the left at 18.0 cm/s. A. Find the velocity of the second marble after the collision.
B. Verify your answer by calculating the total kinetic energy before and after the collision.

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Answer from: graceception

(A) The velocity will be "22.5 m/s".(B) The initial as well as the final K.E will be "6.22×10⁻⁴ J".

Given:

First marble:

Velocity, $u_1 = 22.5 \ m/s$ Mass, $m = m_1 = 0.015 \ kg$

Second marble,

Velocity, $u_2 = -18 \ cm/s$ Mass, $m = m_2 = 0.015 \ kg$

(A)

As we know, the coefficient of restitution will be:

→ $e = \frac{v_2-v_1}{u_1-u_2}$

By putting the values,

$1 = \frac{v_2-(-18)}{22.5-(-18)}$

By applying cross-multiplication,

22.5+18=v_2+18

$v_2 = 22.5 \ m/s$ (velocity)

(B)

The initial K.E:

= $0.5\times 0.015(22.5\times 10^{-2})^2+0.5\times 0.015(18\times 10^{-2})^2$

= $6.22\times 10^{-4 \ J$

The final K.E:

= $0.5\times 0.015(18\times 10^{-2})^2+0.5\times 0.015(22.5\times 10^{-2})^2$

= $6.22\times 10^{-4 \ J$

Thus the responses above is right.

Learn more about collision here:

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Answer from: samiiegarciia

Explanation:

mass of first marble $m_1=m=0.015\ kg$

Initial velocity of the first marble $u_1=22.5\ m/s$

considering right side as positive

Mass of second marble

$m_2=m=0.015\ kg\\u_2=-18\ cm/s$

After collision first marble moves to the left with a velocity of 18 cm/s

i.e. $v_1=-18\ cm/s$

considering v_2 be the velocity of second marble after collision

The Coefficient of restitution is 1 for an elastic collision

$e=\frac{v_2-v_1}{u_1-u_2}$

Putting values

$1=\frac{v_2-(-18)}{22.5-(-18)}\\22.5+18=v_2+18\\v_2=22.5\ m/s$

So, the velocity of the second marble is 22.5 m/s to the right after the collision

(b)Initial kinetic energy = $0.5\times 0.015\times (22.5\times 10^{-2})^2+0.5\times 0.015\times (18\times 10^{-2})^2=6.22\times 10^{-4}\ J$

Final kinetic energy=

$0.5\times 0.015\times (18\times 10^{-2})^2+0.5\times 0.015\times (22.5\times 10^{-2})^2=6.22\times 10^{-4}\ J$

Answer from: Quest

1.the balanced chemical equation is as below

cu + 2 ag(no)₃ → cu(no₃)₂ +2 ag

2.the much silver produced is 50.9 g

step 1: find the moles of cu

moles = mass÷ molar mass

from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236 moles

step 2: use the mole ratio to determine the moles of ag

from equation above cu: ag is 1: 2

therefore the moles of ag =0.236 moles x 2/1 =0.472 moles

step 3: find the mass of ag

mass = moles x molar mass

from periodic table the molar mass of ag =107.87 g/mol

mass = 0.472 moles x 107.87 g/mol = 50.9 g

3. the much silver collected worth is 1.797 oz

convert 50.9 g to oz

that is 1 gram = 0.0353 oz

50.9 grams =? oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1 gram} =1.797 oz

Answer from: Quest

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.

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A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-...