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2 years ago

Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision

Physics

2 answers:

Temka [501]2 years ago

6 0

Explanation:

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look at the attachment above ☝️ and if you have any questions your welcome.

Sidana [21]2 years ago

4 0

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

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We are given a piece of copper of mass m=0.82 g to form a cylindrical wire of resistance R=0.87.2. What should be the length / o

Genrish500 [490]

Answer:

Length of wire=2.16 m

Diameter of wire=0.232 mm

Explanation:

m= mass of copper wire= 0.82 g

R= Resistance of copper wire= 0.87 ohms

D= Density of copper= 8.96 g/cm^3

ρ= Resistivity= 1.7×10^-8 Ωm

$Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\$

$\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m$

$\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm$

3 0

3 years ago

Two small, identical conducting spheres repel each other with a force of 0.030 N when they are 0.65 m apart. After a conducting

solong [7]

Note that the methods applied in solving this question is the appropriate method. Check the parameters you gave in the question if you did not expect a complex number for the charges. Thanks

Answer:

$q_1 = 0.00000119 + j0.00000145 C \\q_2 = 0.00000119 - j0.00000145 C$

Explanation:

Note: When a conducting wire was connected between the spheres, the same charge will flow through the two spheres.

The two charges were 0.65 m apart. i.e. d = 0.65 m

Force, F = 0.030 N

The force or repulsion between the two charges can be calculated using the formula:

$F = \frac{kq^2}{d^2} \\\\0.030 = \frac{9 * 10^9 * q^2}{0.65^2}\\\\q = 1.19 * 10^{-6} C$

Due to the wire connected between the two spheres, $q_1 = q_2 = 1.19 * 10^{-6} C$

The sum of the charges on the two spheres = $q_1 + q_2 = 2.38 * 10^{-6} C$

Note: When the conducting wire is removed, the two spheres will no longer contain similar charges but will rather share the total charge unequally

Let charge in the first sphere = $q_1$

Charge in the second sphere, q₂ = $2.38 * 10^{-6} - q_1$

Force, F = 0.075 N

$F = \frac{k q_1 q_2}{r^2} \\\\0.075 = \frac{9*10^9 * q_1 * (2.38*10^{-6} -q_1 )}{0.65^2}\\\\3.52 * 10^{-12} = q_1 * (2.38*10^{-6} -q_1 )\\\\3.52 * 10^{-12} = 2.38*10^{-6} q_1 - q_1^2\\\\q_1^2 - (2.38*10^{-6}) q_1 + (3.52 * 10^{-12}) = 0\\$