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kifflom [539]
3 years ago
14

A force of 1 N is the only horizontal force exerted on a block, and the horizontal acceleration of the block is

Physics
1 answer:
SVEN [57.7K]3 years ago
4 0

The mass of the first block will be three times the mass of the second block.

According to Newton's second law of motion, the force acting on a body is directly proportional to the acceleration as shown;

F\ \alpha \ a

F = ma

F is the acting force

m is the mass

a is the acceleration of the body

Given the following parameters

Constant force F =  1N

For the first block with the acceleration of "a"

1 = m₁a

a = m₁/1

m₁ = a .................1

For the second block, acceleration is thrice that of the first. This means;

F = m(3a)

1 = 3ma

m_2=\frac{1}{3a} ..........................2

Divide both equations

\frac{m_1}{m_2} =\frac{a}{(\frac{1}{3a} )}\\\frac{m_1}{m_2} = 3\\m_1 = 3m_2

From the calculation, we can conclude that the mass of the first block will be three times the mass of the second block.

Learn more here: brainly.com/question/19030143

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Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

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