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2 years ago

What can you conclude about the velocity of the object on the graph below

Physics

1 answer:

Mars2501 [29]2 years ago

7 0

Answer:

The more time goes on the higher in meters. So most likely more time = closer to terminal velocity. -hope this helps.

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$