Question

Particle A of mass 1 kg is moving with velocity 2 m/s along positive x-axis. Particle B of mass 3 kg is moving with velocity 4 m/s along positive y-axis. At time t = 0, the co-ordinates (in metre) of A and B are (3, 0) and (0, 8) respectively. The plane is horizontal and coefficient of friction is 0.2. The value of y co-ordinate of centre of mass (in metre), when centre of mass will finally stop $(g=10m/s^2)$.___

Answer 1

Particle A will stop at
$x_1=3+\frac{v_1^2}{2\mu g}=3+\frac{(2{)}^2}{2\left(0.2\right)\left(10\right)}=4m,x_2=0$
Particle B will stop at
$y_1=0,y_1=8+\frac{v_2^2}{2\mu g}=8+\frac{(4{)}^2}{2\left(0.2\right)\left(10\right)}=12m$
y co-ordinate of centre of mass
$y_{\infty m}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{\left(1\right)\left(0\right)+3\left(12\right)}{4}=9m$