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Evgen [1.6K]
3 years ago
13

g A lighter ball A with momentum PA=5.0 kg∙m/s in positive x-direction approaches a heavier ball B at rest before the collision.

After this head-on collision, the lighter ball bounces straight back with a momentum P’A=2.0 kg∙ m/s in negative x-direction. What is the magnitude of the heavier ball B’s momentum change?
Physics
1 answer:
Svet_ta [14]3 years ago
3 0

Answer:

7kgm/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

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A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there
jenyasd209 [6]

Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

Area of crossection, A = 13 cm²

Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

e=\frac{NBA}{t}

Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

4 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the c
PtichkaEL [24]

Answer:

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2

3 0
3 years ago
Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher
siniylev [52]

Answer:

v = 4.44 \times 10^5 m/s

Explanation:

By Einstein's Equation of photoelectric effect we know that

h\nu = W + \frac{1}{2}mv^2

here we know that

h\nu = energy of the photons incident on the metal

W = minimum energy required to remove photons from metal

\frac{1}{2}mv^2 = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

W = \frac{hc}{351 nm}

here we know that

hc = 1242 eV-nm

now we have

W = \frac{1242}{351} = 3.54 eV

now by energy equation above when photon of 303 nm incident on the surface

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4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2

(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2

8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2

v = 4.44 \times 10^5 m/s

6 0
3 years ago
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