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Question Number 190902 by Spillover last updated on 13/Apr/23

Answered by mr W last updated on 13/Apr/23

$$\left(i\right)$$$$(m_A+m_B)a=(m_B-m_A)g$$$$\Rightarrow a=\frac{(m_B-m_A)g}{m_B+m_A}=\frac{(0.7-0.3)\times 10}{0.7+0.3}=4m/s^2$$$$T=m_B(g-a)=m_B(1-\frac{m_B-m_A}{m_B+m_A})g=\frac{2m_A{m}_{B}g}{m_A+m_B}$$$$=\frac{2\times 0.3\times 0.7\times 10}{0.3+0.7}=4.2N✓$$$$\left(ii\right)$$$$v=at=1.6$$$$\Rightarrow t=\frac{1.6}{4}=0.4s$$$$h=\frac{{at}^2}{2}=\frac{4\times 0.4^2}{2}=0.32m$$$$0.32+0.52=-1.6t+\frac{{gt}^2}{2}$$$$5t^2-1.6t-0.84=0$$$$\Rightarrow t=\frac{1.6+\sqrt{1.6^2+4\times 5\times 0.84}}{2\times 5}=0.6s✓$$

Commented by Spillover last updated on 14/Apr/23

$$nicesolution.thanks$$

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