Question

Two particles A and $B$ of mass $1kg$ and $2kg$ respectively are projected in the directions shown in figure with speeds $u_A=200m/s$ and $u_B=50m/s$, Initially they were $90m$ apart. Find the maximum height attained by the centre of mass of the particles. $(g=10m/s^2)$

• A. $115.55m$
• B. $145.55m$
• C. $4.55m$
• D. $34.55m$

Answer 1

Answer: (A)

$115.55m$

Now, Height of centre of mass $=\frac{m_1{H}_1+m_2{H}_2}{m_1+m_2}$

$=\frac{200\times 0+90\times 2}{(2+1)}=\frac{180}{3}=60m$

Now velocity of centre of mass $=\frac{V_1{m}_1+V_2{m}_2}{m_1+m_2}$

$=\frac{200\times 1-50\times 2}{2+1}=\frac{100}{3}m/s$

$V^2=u^2+2as\Rightarrow O^2={\left(\frac{100}{3}\right)}^2-2\times 10\times h$ [final velocity $=0$ to get maximum height]

$\Rightarrow h_1=\frac{100\times 100}{9\times 2\times 10}=55.55m$

Total height obtained by center of mass $=(60+55.55)m=115.55m$

Correct option is (A)