Photo AI
4. (a) Two particles of masses 0.24 kg and 0.25 kg are connected by a light inextensible string passing over a small, smooth, fixed pulley - Leaving Cert Applied Maths - Question 4 - 2010
Question 4
4. (a) Two particles of masses 0.24 kg and 0.25 kg are connected by a light inextensible string passing over a small, smooth, fixed pulley. The system is released f... show full transcript
Worked Solution & Example Answer:4. (a) Two particles of masses 0.24 kg and 0.25 kg are connected by a light inextensible string passing over a small, smooth, fixed pulley - Leaving Cert Applied Maths - Question 4 - 2010
Step 1
(i) the tension in the string
Answer
To determine the tension in the string, we can apply Newton's second law.
Let:
-
For the 0.24 kg mass:
-
The weight = 0.24g, where g = 9.81 m/s². The resultant force is given by:
a = 0.25g - T = 0.25f
0.24g - T = 0.24f
Combining these equations, we find that:
T = 2.4 N.
Step 2
(ii) the speed of the two masses when the 0.25 kg mass has descended 1.6 m
Answer
To find the speed of the masses when the 0.25 kg mass has descended 1.6 m, we can use the equation of motion:
v² = u² + 2as
Where:
- u = initial velocity = 0 (since released from rest)
- a = acceleration (which we found from part (i)) = 0.2 m/s²
- s = distance = 1.6 m
Substituting the values:
v² = 0 + 2(0.2)(1.6)
Solving this gives us:
v = ext{√(0.64)} = 0.8 m/s.
Step 3
(i) Show, on separate diagrams, the forces acting on the wedge and on the particles.
Answer
For the wedge:
- The normal force (R) acts vertically upwards.
- The weight of the wedge (4mg) acts downwards.
- The frictional force (S) acts parallel to the surface.
For the particles:
- For 2m: The forces are the downward gravitational force (2mg) and tension (T) acting along the string.
- For m: The forces are similar, including the gravitational force (mg) and tension (T).
Step 4
(ii) Find the acceleration of the wedge.
Answer
Using the equations of motion, we have:
For the wedge:
2m * 2mg * cos(45°) - R = 2mf sin(45°)
And, R = √(2(mg - mf))
For the small particle:
mg cos(45°) - S = m * f * sin(45°)
This leads to:
S = 1/√2 * (mg - mf)
Applying conservation of momentum gives:
4m * S sin(45°) + R sin(45°) = 4m.
Finally, this leads to the acceleration f = 3g/11, or approximately 2.67 m/s².