All categories

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

Physics

1 answer:

arlik [135]3 years ago

3 0

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

You might be interested in

A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

garik1379 [7]

U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2

6 0

3 years ago

Read 2 more answers

When melting of a metamorphic rock occurs, it changes into what?

Thepotemich [5.8K]

Answer:

The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.

Explanation:

i hope this helps

8 0

2 years ago

A displacement vector 5 units long makes an angle of 67° with the x-axis. What is the horizontal component of displacement?

Harman [31]

Yo sup??

magnitude of original vector is 5 units.

angle made with x axis is 67°

horizontal component=5*cos67°

=5*0.4

=2 units (approx)

Therefore the correct answer is option B

Hope this helps

7 0

3 years ago

Read 2 more answers

a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am

Sindrei [870]

Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

θ = angle between wire's current direction and magnetic field = 90°

0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

3 0

3 years ago

Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

3 years ago