Q3.2 Forces and accelerations (H)
A certain force when applied to a mass m1 gives it an acceleration of 4 ms-2. When appled to a mass m2 it gives this mass an acceleration of 3 ms-2. Find the acceleration that the force would give:- a body of mass m2+m1
- a body of mass m2-m1
Hint
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Hint
You’ll need to write the 2nd Law four times, each time with the same force. And don’t substitute numbers until you have to!Solution
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The ratio of the two masses follows as
![\[ \frac{m_2}{m_1} =\frac{a_1}{a_2} = \frac{4}{3} \]](mastermathpng-1.png)
If the force is applied to a body of mass m2-m1 the acceleration (let’s call
it a-) satisfies

Solution
Let a1=4 ms-2 and a2=3 ms-2 be the accelerations of m1 and m2 when subjected separately to the unknown force, F. Then Newton’s 2nd Law applied to each of these cases implies
![\[ F= m_1 a_1 \hspace{1cm}\mbox{\rm and}\hspace{1cm} F= m_2 a_2 \]](mastermathpng-0.png)
![\[ \frac{m_2}{m_1} =\frac{a_1}{a_2} = \frac{4}{3} \]](mastermathpng-1.png)
Let a+ be the acceleration of the body of mass m1+m2. Then
F=(m1+m2)a+
Combining this equation with the first of the opening pair of equations gives
(m1+m2)a+=m1a1
from which we find
![\[ a_+ = \frac{m_1a_1}{m_1+m_2} = \frac{a_1}{1+m_2/m_1} =\frac{a_1}{1+4/3} = \frac{3}{7}a_1 = \frac{12}{7} ms^{-2} \]](mastermathpng-2.png)
F=(m2-m1)a-
Appealing again to the opening pair of equations
(m2-m1)a-=m1a1
implying
![\[ a_- = \frac{m_1a_1}{m_2-m_1} = \frac{a_1}{4/3-1} = 3a_1 =12\, ms^{-2} \]](mastermathpng-3.png)