Two particles of masses 1 kg and 3 kg have position vectors 2 -3 -4 k- and -2 -3 -4 k- respectively- The centre of mass has a position vector-A- -3 -2 k-B- -3 -2 k-C- -3 -2 k-D- -3 -2 k-

The correct option is D î+3ĵ 2k̂ Position vector of centre of mass r_cm=m_1 r_1+m_2 r_2/m 1 +m_2 r_1 =2î+3ĵ+4k̂, m_1 =1kg r_2 = 2î+3ĵ 4k̂, m_2 =3kg r_cm=1( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Center of Mass as an Average Point

Two particles...

Question

Two particles of masses 1kg and 3kg have position vectors $2^i + 3^j + 4^k$ and $- 2^i + 3^j - 4^k$ respectively. The centre of mass has a position vector:

$^i + 3^j - 2^k$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$-^i - 3^j - 2^k$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$-^i + 3^j + 2^k$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$-^i + 3^j - 2^k$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$-^i + 3^j - 2^k$

Position vector of centre of mass
$r_{c m} = \frac{m_{1} r_{1} + m_{2} r_{2}}{m - 1 + m_{2}} r_{1} = 2^i + 3^j + 4^k, m_{1} = 1 k g r_{2} = - 2^i + 3^j - 4^k, m_{2} = 3 k g r_{c m} = \frac{1 (2^i + 3^j + 4^k) + 3 (- 2^i + 3^j - 4^k)}{1 + 3} = \frac{4^i + 12^j - 8^k}{4} = - i + 3 j - 2^k$

Suggest Corrections

Similar questions

Two particles of mass 1 kg and 3 kg have position vectors $2^i + 3^j + 4^k a n d - 2^i + 3^j - 4^k$ respectively. The centre of mass has a position vector