Physics, 15.07.2020 17:01 ella5916

A 30.0-g object moving to the right at 19.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.) 30.0-g object Incorrect: Your answer is incorrect. seenKey 16.8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm/s 13.0-g object

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Answer from: pinkpearl2022

The final velocity of the 30 g object is 16.8 cm/s

The final velocity of the 13 g object is 21.3 cm/s

Explanation:

Let's study the elastic collision with conservation of linear momentum, assigning object 1 to the 30 g object, and object 2 to the 13 gr object:

$p_{1\,i}+p_{2\,i}=p_{1\.f}+p_{2\,f}\\(30)\,(19.5)+(13)\,(15) = (30)\,v_{1\,f} +(13)\,v_{2\,f}\\780 = (30)\,v_{1\,f} +(13)\,v_{2\,f}$

so we can write one of the unknowns in terms of the other one:

$v_{1\,f}=(780-13\,v_{2\,f})/30$

Now we analyze the equation for conservation of kinetic energy that verifies in elastic collisions:

$\frac{30}{2} (19.5)^2+\frac{13}{2} \,(15)^2=\frac{30}{2} (v_{1\,f})^2+\frac{13}{2} \,(v_{2\,f})^2\\7166.25=15\, (v_{1\,f})^2+6.5\,(v_{2\,f})^2$

now we can write this quadratic equation replacing $v_{1\,f}$ with its expression in terms of $v_{2\,f}$ and solve it (with the help of a graphing calculator is simpler by looking for the roots).

We get two answers for $v_{2\,f}$ : one 15 cm/s, and the other one 21.28 cm/s.

We select the 21.28 cm/s answer since otherwise, the situation is the same as the initial one at which the second object was moving at 15 cm/s.

This velocity can be rounded to one decimal to: 21.3 cm/s

Given the value 21,28 for $v_{2\,f}$ , then:

$v_{1\,f}=(780-13\,v_{2\,f})/30\\v_{1\,f}=(780-13\,(21.28))/30=16.78 \,\,cm/s$

which can be rounded to 16.8 cm/s

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

Iam looking for this answer too!

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