A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1
Question 3
A particle P of mass 2 kg is moving under the action of a constant force F newtons. When t = 0, P has velocity (3i + 2j) m s⁻¹ and at time t = 4 s, P has velocity (1... show full transcript
Worked Solution & Example Answer:A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1
Step 1
the acceleration of P in terms of i and j
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Answer
To find the acceleration of P, we first determine the change in velocity over the time interval. The initial velocity at t = 0 s is ( (3i + 2j) ) m/s and the velocity at t = 4 s is ( (15i - 4j) ) m/s.
The change in velocity, ( \Delta v ), can be calculated as:
Δv=(15i−4j)−(3i+2j)=(15i−3i)+(−4j−2j)=12i−6j
The time interval, ( \Delta t ), is 4 s. Thus, the acceleration ( a ) is given by the formula:
a=ΔtΔv=412i−6j=3i−1.5j m/s2
Step 2
the magnitude of F
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Answer
Using Newton's second law, we have:
F=ma
where ( m = 2 ) kg and we found ( a = (3i - 1.5j) ) m/s².
Substituting the acceleration:
F=2(3i−1.5j)=6i−3j N
To find the magnitude of the force F:
∣F∣=(6)2+(−3)2=36+9=45≈6.71 N
Step 3
the velocity of P at time t = 6 s
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Answer
To find the velocity of P at time t = 6 s, we can use the acceleration we calculated previously.
Starting with the velocity at t = 4 s and applying the acceleration over an additional 2 seconds:
Initial velocity at t = 4 s: ( (15i - 4j) ) m/s
Using the formula:
v=u+at
where ( u = (15i - 4j) ) m/s, ( a = (3i - 1.5j) ) m/s², and ( t = 2 ) s:
v=(15i−4j)+(3i−1.5j)(2)=(15i−4j)+(6i−3j)=(15+6)i+(−4−3)j=21i−7j m/s
