Three masses are placed on the x-axis: \(300\) g at the origin, \(500\) g at \(x =40\) cm, and \(400\) g at \(x=70\) cm. The distance of the center of mass from the origin is:
1. | \(40\) cm | 2. | \(45\) cm |
3. | \(50\) cm | 4. | \(30\) cm |
A uniform square plate \(ABCD\) has a mass of \(10\) kg.
If two point masses of \(5\) kg each are placed at the corners \(C\) and \(D\) as shown in the adjoining figure, then the centre of mass shifts to the mid-point of:
1. \(OH\)
2. \(DH\)
3. \(OG\)
4. \(OF\)
The mass per unit length of a non-uniform rod of length \(L\) is given by \(\mu =λx^{2}\) where \(\lambda\) is a constant and \(x\) is the distance from one end of the rod. The distance between the centre of mass of the rod and this end is:
1. | \(\frac{L}{2}\) | 2. | \(\frac{L}{4}\) |
3. | \(\frac{3L}{4}\) | 4. | \(\frac{L}{3}\) |
The centre of the mass of \(3\) particles, \(10~\text{kg},\) \(20~\text{kg},\) and \(30~\text{kg},\) is at \((0,0,0).\) Where should a particle with a mass of \(40~\text{kg}\) be placed so that its combined centre of mass is \((3,3,3)?\)
1. \((0,0,0)\)
2. \((7.5, 7.5, 7.5)\)
3. \((1,2,3)\)
4. \((4,4,4)\)
Two particles of mass, \(2\) kg and \(4\) kg, are projected from the top of a tower simultaneously, such that \(2\) kg of mass is projected with a speed \(20\) m/s at an angle \(30^{\circ}\) above horizontal and \(4\) kg is projected at \(40\) m/s horizontally. The acceleration of the centre of mass of the system of two particles will be:
1. \(\dfrac{g}{2}\)
2. \(\dfrac{g}{4}\)
3. \(g\)
4. \(2g\)
Five uniform circular plates, each of diameter \(D\) and mass \(m,\) are laid out in a pattern shown. Using the origin shown, the \(y\text-\text{coordinate}\) of the centre of mass of the ''five–plate'' system will be:
1. | \(\frac{2D}{5}\) | 2. | \(\frac{4D}{5}\) |
3. | \(\frac{D}{3}\) | 4. | \(\frac{D}{5}\) |
1. | \(9.9~\text m\) | 2. | \(10.1~\text m\) |
3. | \(10~\text m\) | 4. | \(20~\text m\) |
At \(t=0,\) the positions of the two blocks are shown. There is no external force acting on the system. Find the coordinates of the centre of mass of the system (in SI units) at \(t=3\) seconds.
1. | \((1,0)\) | 2. | \((3,0)\) |
3. | \((4.5,0)\) | 4. | \((2.25,0)\) |
A bomb is projected from the ground at a horizontal range of \(R\). If the bomb explodes mid-air, then the range of its centre of mass is:
1. \(\frac{R}{2}\)
2. \(R\)
3. \(2R\)
4. \(\frac{2R}{3}\)