`x-y` is the vertical plane as shown in figure. A particle of mass `1kg` is at `(10m, 20m)` at time `t=0`. It is released from rest. Another particles of mass `2kg` is at `(20m, 40m)` at the same instant. It is projected with velocity `(10hati+10hatj)m//s`. After `1s`. Find (a) acceleration (b)velocity (c) position of the center of mass

Answer

A, B, C

Explanation

(a) `a_(COM)=(m_1a_1+m_2a_2)/(m_1+m_2)`
`=((1)(-10hatj)+(2)(-10hatj))/(1+2)`
`=(-10hatj)m//s^2`
(b) `v_(COM)=(m_1v_1+m_2v_2)/(m_1+m_2)`
`=(m_1(u_1+a_1t)+m_2(u_2+a_2t))/(m_1+m_2)`
`=((1)[0+(-10hatj)(1)]+2[(10hati+10hatj)+(-10hatj)(1)])/(1+2)`
`=10/3(2hati-hatj)m//s`
(c) `r_(CM)=(m_1r_1+m_2r_2)/(m_1+m_2)`
`=(m_1[r_1+ut+1/2at^2]_1+m_2[r_1+ut+1/2at^2]_2)/(m_1+m_2)`
`=(1[(10hati+20hatj)+0+1/2(-10hatj)(1)^2]+2[(20hati+40hatj)+(10hati+10hatj)(1)+1/2(-10hatj)(1)^2])/(1+2)`
`=(70/3hati+35hatj)m`