Two forces F_1 and F_2 act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1
Question 7
Two forces F_1 and F_2 act on a particle P.
The force F_1 is given by F_1 = (-i + 2j) N and F_2 acts in the direction of the vector (i + j).
Given that the resulta... show full transcript
Worked Solution & Example Answer:Two forces F_1 and F_2 act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1
Step 1
find F_2
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Answer
To find the force F_2, we start by letting it be represented as:
F2=k(i+j)
where k is a scalar that represents the magnitude of the force. We know that the resultant of F_1 and F_2 acts in the direction of the vector (i+3j).
Finding the resultant vector:
The resultant force can be expressed as:
FR=F1+F2=(−i+2j)+k(i+j)
This combines to:
FR=(−1+k)i+(2+k)j
Setting up the equation for direction:
Since FR must act in the direction of (i+3j), we can set up a ratio:
rac{-1 + k}{1} = rac{2 + k}{3}
Cross-multiplying to solve for k:
-1 + k = rac{1}{3}(2 + k)
Multiplying through by 3 gives:
3(−1+k)=2+k
Which simplifies to:
−3+3k=2+k
Rearranging yields:
ightarrow k = 2.5 $$
Substituting back to find F2:
Therefore, the force F2 is:
F2=2.5(i+j)=(2.5i+2.5j)extN
Step 2
Find the speed of P when t = 3 seconds
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Answer
The acceleration of P is given by:
a=(3i+9j)extms−2
Using kinematic equations:
The velocity of P at any time t can be found using the formula:
v=u+at
where:
u is the initial velocity: u=(3i−22j)extms−1
a is the acceleration: a=(3i+9j)extms−2
t=3 seconds
Substituting the values:
v=(3i−22j)+(3i+9j)(3)
Simplifying this:
v=(3i−22j)+(9i+27j)v=(12i+5j)extms−1
Calculating the speed:
The speed is the magnitude of the velocity vector:
∣v∣=sqrt(12)2+(5)2=sqrt144+25=sqrt169=13extm/s
Thus, the speed of P when t=3 seconds is 13extm/s.
