Physics, 21.04.2020 15:51 gwendallinesikes

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the collision is half its speed before the collision. Block B was at rest before the collision. The mass of block A is 7 kg and the mass of block B is 2 kg. What is the minimum initial speed (in m/s) that block A must have for block B to swing through a complete vertical circle?

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Answer from: browneyedbaby20

Complete Question

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

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

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

making v_A the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

v_A= 4m/s

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