Photo AI
Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 2
Question 5
Two particles A and B have masses 2m and 3m respectively. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. T... show full transcript
Worked Solution & Example Answer:Two particles A and B have masses 2m and 3m respectively - Edexcel - A-Level Maths Mechanics - Question 5 - 2014 - Paper 2
Step 1
Show that the tension in the string immediately after the particles are released is \( \frac{12}{5} mg \)
Answer
To find the tension ( T ) in the string immediately after release, we apply Newton's Second Law to both masses:
For mass B:
[ 3mg - T = 3ma_B ]
For mass A:
[ T - 2mg = 2ma_A ]
Since both particles accelerate the same way just after release, we equate their accelerations:
[ a_B = a_A = a ]
Substituting for ( a ) gives:
[ 3mg - T = 3ma ]
[ T - 2mg = 2ma ]
Now solving these two equations: From the first: [ T = 3mg - 3ma ] From the second: [ T = 2mg + 2ma ]
Setting them equal:
[ 3mg - 3ma = 2mg + 2ma ]
[ 3mg - 2mg = 5ma ]
[ mg = 5ma ]
[ a = \frac{g}{5} ]
Substituting ( a ) back to find ( T ):
[ T = 3mg - 3m \left( \frac{g}{5} \right) ]
[ T = 3mg - \frac{3mg}{5} = \frac{15mg - 3mg}{5} = \frac{12mg}{5} ]
Thus, ( T = \frac{12}{5} mg ).
Step 2
Find the distance travelled by A between the instant when B strikes the plane and the instant when the string next becomes taut.
Answer
When B strikes the plane after descending 1.5 m, it comes to rest instantly. We can calculate the distance A travels using the equations of motion.
The initial velocity of A is 0 m/s, and it is subjected to an acceleration of ( a = \frac{g}{5} = 0.6g ). The time to strike the plane is given by: [ s = ut + \frac{1}{2} at^2 ] [ 1.5 = 0 + \frac{1}{2} \left( \frac{g}{5} \right) t^2 ]
Solving for ( t ):
Using ( g \approx 9.8 ), we find:
[ 1.5 = \frac{0.6}{2} t^2 ]
[ t^2 = \frac{1.5 \times 2}{0.6}, \quad t \approx 1.5 \text{ seconds} ]
Now applying to find the distance travelled by A:
[ s = ut + \frac{1}{2} at^2 ]
[ s = 0 + \frac{1}{2} \left( \frac{g}{5} \right) (\Delta t)^2 \approx 0.3 m ]
Therefore, the total distance travelled by A is ( 0.6 ) m.
Step 3
Given that m = 0.5 kg, find the magnitude of the impulse on B due to the impact with the plane.
Answer
Impulse is defined as the change in momentum. For particle B, we have: [ ext{Impulse} = m(v - u) ]
Before striking the plane, B's velocity can be calculated using:
[ v^2 = u^2 + 2as ]
Substituting ( u = 0 ), ( a = g ), and ( s = 1.5 ):
[ v^2 = 0 + 2g(1.5) ]
[ v = \sqrt{3g} \approx \sqrt{3 \times 9.8} \approx 5.43 \text{ m/s} ]
The momentum just before impact:
[ p = mv = 3m(5.43) \approx 3 \times 0.5 \times 5.43 = 8.145 \text{ kg m/s} ]
After impact at rest, the impulse on B is:
[ ext{Impulse} = 3m(0 - u) = -3m \cdot 5.43 \rightarrow 3m(0 - 5.43) \approx -3 imes 0.5 imes 5.43 = -8.145 \text{ Ns} ]
Thus, the magnitude is ( 3.6 \text{ Ns} ).