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A small ball, P.of mass 0.8kg, is held at rest on a smooth horizontal table and is attached to one end of a thin rope. The rope passes over a pulley that is fixed at the edge of the table. The other end of the rope is attached to another small ball , Q, of mass 0.6kg , that hangs freely below the pulley. Ball P is released from rest, with the rope taut, with P at a distance of 1.5 m from the pulley and with Q at a height of 0.4m above the horizontal floor, as shown in Figure 1. Ball Q descends, hits the floor and does not rebound. The balls are modelled as particles, the rope as a light and inextensible string and the pulley as small and smooth. Using this model, (a) show that the acceleration of Q, as it falls, is 4.2ms^-2 (5) (b) find the time taken by P to hit the pulley from the instant when P is released. (6) (c) State one limitation of the model that will affect the accuracy of your answer to part (a). (1) Total (12 marks)

Question

A small ball, P.of mass 0.8kg, is held at rest on a smooth horizontal table and is attached to one end of a thin rope. The rope passes over a pulley that is fixed at the edge of the table. The other end of the rope is attached to another small ball , Q, of mass 0.6kg , that hangs freely below the pulley. Ball P is released from rest, with the rope taut, with P at a distance of 1.5 m from the pulley and with Q at a height of 0.4m above the horizontal floor, as shown in Figure 1. Ball Q descends, hits the floor and does not rebound. The balls are modelled as particles, the rope as a light and inextensible string and the pulley as small and smooth. Using this model, (a) show that the acceleration of Q, as it falls, is 4.2ms^-2 (5) (b) find the time taken by P to hit the pulley from the instant when P is released. (6) (c) State one limitation of the model that will affect the accuracy of your answer to part (a). (1) Total (12 marks)

A small ball, P.of mass 0.8kg, is held at rest on a smooth horizontal table and is attached to one end of a thin rope. The rope passes over a pulley that is fixed at the edge of the table. The other end of the rope is attached to another small ball , Q, of mass 0.6kg , that hangs freely below the pulley. Ball P is released from rest, with the rope taut, with P at a distance of 1.5 m from the pulley and with Q at a height of 0.4m above the horizontal floor, as shown in Figure 1. Ball Q descends, hits the floor and does not rebound. The balls are modelled as particles, the rope as a light and inextensible string and the pulley as small and smooth. Using this model, (a) show that the acceleration of Q, as it falls, is 4.2ms^-2 (5) (b) find the time taken by P to hit the pulley from the instant when P is released. (6) (c) State one limitation of the model that will affect the accuracy of your answer to part (a). (1) Total (12 marks)

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WestonElite · Tutor for 8 years

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(a) The acceleration of \( Q \), as it falls, is \( 4.2 \mathrm{~ms}^{-2} \).<br /><br />(b) The time taken by \( P \) to hit the pulley is calculated using the acceleration from part (a) and the distance \( P \) has to travel.<br /><br />(c) One limitation of the model is the assumption of a frictionless pulley. In reality, friction in the pulley would affect the tension in the rope and the resulting acceleration and time.

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## Step 1: Determine the acceleration of ball Q.<br />Since the rope is inextensible and light, the tension in the rope is the same throughout its length. The only forces acting on ball Q are the tension \(T \) and the weight \( W_Q = m_Qg \), where \( m_Q \) is the mass of ball Q and \( g \) is the acceleration due to gravity (approximately \( 9.8 \mathrm{~ms}^{-2} \)). The net force on ball Q is \( W_Q - T \).<br /><br />### \(\boldsymbol{F_{net,Q} = m_Qg - T}\)<br /><br />## Step 2: Determine the acceleration of ball P.<br />For ball P, the only horizontal force is the tension \(T \) in the rope. Since there are no other horizontal forces, the net force on ball P is \( T \).<br /><br />### \(\boldsymbol{F_{net,P} = T}\)<br /><br />## Step 3: Apply Newton's second law to both balls.<br />For ball Q, Newton's second law gives us:<br /><br />### \(\boldsymbol{m_Qg - T = m_Qa}\)<br /><br />For ball P, Newton's second law gives us:<br /><br />### \(\boldsymbol{T = m_Pa}\)<br /><br />## Step 4: Combine the equations to solve for acceleration \(a \).<br />Adding the two equations eliminates \( T \):<br /><br />### \(\boldsymbol{m_Qg = (m_Q + m_P)a}\)<br /><br />Now, we can solve for \(a \):<br /><br />### \(\boldsymbol{a = \frac{m_Qg}{m_Q + m_P}}\)<br /><br />## Step 5: Calculate the acceleration \(a \).<br />Substitute the given values \( m_Q = 0.6 \mathrm{~kg} \), \( m_P = 0.8 \mathrm{~kg} \), and \( g = 9.8 \mathrm{~ms}^{-2} \) into the equation for \( a \).<br /><br />## Step 6: Find the time taken by \( P \) to hit the pulley.<br />Using the equation of motion \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time, we can solve for \( t \) since \( u = 0 \) (ball P is released from rest), \( s = 1.5 \mathrm{~m} \), and \( a \) is the acceleration we found in step 5.<br /><br />### \(\boldsymbol{s = \frac{1}{2}at^2}\)<br /><br />## Step 7: Solve for time \(t \).<br />Rearrange the equation to solve for \( t \):<br /><br />### \(\boldsymbol{t = \sqrt{\frac{2s}{a}}}\)<br /><br />## Step 8: Calculate the time \( t \).<br />Substitute the values of \( s = 1.5 \mathrm{~m} \) and \( a \) from step 5 into the equation for \( t \).<br /><br />## Step 9: Discuss the limitation of the model.<br />One limitation of the model is the assumption that the pulley is small and smooth, which means there is no friction. In reality, there would be some friction in the pulley, which would affect the tension in the rope and thus the acceleration and time calculated.
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