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deff fn [24]
2 years ago
13

A 15 kg block is sliding along a frictionless surface and strikes a 10 kg ball at rest. What is the collision of the blocks afte

r the collision if it is an inelastic collision?
Physics
1 answer:
vfiekz [6]2 years ago
7 0

Answer:

   v = 0.6 v₁

Explanation:

This is an exercise in collisions, let's start by defining a system formed by the two bodies, so that the forces during the collisions have been internal and the momentum is preserved.

Instant starts. Before the crash

       p₀ = M v₁ + m 0

Final moment. After the crash

       p_{f} = (M + m) v

how momentum is conserved

       p₀ = p_{f}

       M v₁ = (M + m) v

       

       v = \frac{M}{M+m} v_{1}

let's calculate

       v = \frac{15}{15+10} v_{1}

        v = 0.6 v₁

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Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
2 years ago
What is the mass of 3 m3 of a substance having density 1200 kg/m3​
adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0
2 years ago
What is 1.0 x 10^9 in standard form?
jasenka [17]
1.0 x 10^9= 1000000000
3 0
3 years ago
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an
Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0
3 years ago
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Brrunno [24]

Answer:D

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