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3 years ago

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A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

ing a perfectly elastic collision, what will be the speed and direction of each object after the collision?

1 answer:

Gekata [30.6K]3 years ago

3 0

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

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(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

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which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

$\rho = \frac{m}{V}$

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

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A car moved 60 meters west in 2 hours. What is its average velocity?

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v=Δx/t

$v=\frac{xf-xi}{t}= \frac{60m}{2}=30m$

<em><u>Final answer is 30.</u></em>

Hope this helps!

Thanks!

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Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

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In this problem, we are given the following:

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Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

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