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bezimeni [28]
3 years ago
10

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

nitially at rest. After the collision, block A continues to the right at 1.0 m/s. The mass of block A is 2.0 kg and the mass of block B is 3.0 kg. The positive x-direction is to the right. a. Find the final velocity of block B. Show your work. b. Is the sign of the change in momentum of block A as a result of the collision positive, negative, or is it zero
Physics
1 answer:
ad-work [718]3 years ago
6 0

Answer: 2.67 m/s

Explanation:

Given

Mass of block A  is m_a=2\ kg

mass of block B is m_b=3\ kg

The initial velocity of block A u_a=5\ m/s

the initial velocity of block B is u_b=0

After collision velocity of block A is v_a=1\ m/s

Conserving momentum

m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s

The momentum of block A after the collision is P_a=2\times 1=2\ kg.m/s

Therefore, there is no change in sign.

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If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st
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Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

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coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

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0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

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Angular momentum is equivalent to:

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Plug in the given values for moment of inertia and angular speed:

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