163791
Two bodies of masses \(\mathbf{5} \mathbf{~ k g}\) and \(\mathbf{2} \mathbf{~ k g}\) are moving with velocity \((\hat{i}+2 \hat{j}-2 \hat{k})\) and \((-2 \hat{i}-5 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) respectively. What is the velocity of the centre of mass:
163792
The position of centre of mass of a system consisting of two particles of masses \(m_1\) and \(m_2\) separated by a distance \(L\) apart, from \(m_1\) will be:
1 \(\frac{m_1 L}{m_1+m_2}\)
2 \(\frac{m_2 L}{m_1+m_2}\)
3 \(\frac{m_2}{m_1} L\)
4 \(\frac{L}{2}\)
Explanation:
it shows from the figure that \( X_{c m}=\frac{m_1 \times 0+m_2 L}{m_1+m_2}=\frac{m_2}{m_1+m_2} L \) \( Y_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) \( Z_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) centre of mass is at a distance \(\left\) from \(m_1\) internally on the line joining the two particles.
NCERT-XI-I -98
4 RBTS PAPER
163793
If the system is released, then acceleration of the center of mass of the system is :
1 \(g / 4\)
2 \(g / 2\)
3 \(\mathrm{g}\)
4 \(2 \mathrm{~g}\)
Explanation:
Acceleration of each mass \( a=\left(\frac{3 m-m}{3 m+m}\right) g=g / 2 \) \( a c c^n \text { of CM, } a_{c m}=\frac{m_1 a_1+m_2 a_2}{m_1+m_2} \) \( a_{c m}=\frac{3 m \frac{g}{2}-m \frac{g}{2}}{3 m+m}=\frac{g}{4} \)
NCERT-XI- I -99
4 RBTS PAPER
163794
Four bodies of masses 2, 3,5 and \(8 \mathrm{~kg}\) are placed at the four corners of a square of side \(2 \mathrm{~m}\). The position of \(\mathrm{CM}\) will be :
NEET Test Series from KOTA - 10 Papers In MS WORD
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4 RBTS PAPER
163791
Two bodies of masses \(\mathbf{5} \mathbf{~ k g}\) and \(\mathbf{2} \mathbf{~ k g}\) are moving with velocity \((\hat{i}+2 \hat{j}-2 \hat{k})\) and \((-2 \hat{i}-5 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) respectively. What is the velocity of the centre of mass:
163792
The position of centre of mass of a system consisting of two particles of masses \(m_1\) and \(m_2\) separated by a distance \(L\) apart, from \(m_1\) will be:
1 \(\frac{m_1 L}{m_1+m_2}\)
2 \(\frac{m_2 L}{m_1+m_2}\)
3 \(\frac{m_2}{m_1} L\)
4 \(\frac{L}{2}\)
Explanation:
it shows from the figure that \( X_{c m}=\frac{m_1 \times 0+m_2 L}{m_1+m_2}=\frac{m_2}{m_1+m_2} L \) \( Y_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) \( Z_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) centre of mass is at a distance \(\left\) from \(m_1\) internally on the line joining the two particles.
NCERT-XI-I -98
4 RBTS PAPER
163793
If the system is released, then acceleration of the center of mass of the system is :
1 \(g / 4\)
2 \(g / 2\)
3 \(\mathrm{g}\)
4 \(2 \mathrm{~g}\)
Explanation:
Acceleration of each mass \( a=\left(\frac{3 m-m}{3 m+m}\right) g=g / 2 \) \( a c c^n \text { of CM, } a_{c m}=\frac{m_1 a_1+m_2 a_2}{m_1+m_2} \) \( a_{c m}=\frac{3 m \frac{g}{2}-m \frac{g}{2}}{3 m+m}=\frac{g}{4} \)
NCERT-XI- I -99
4 RBTS PAPER
163794
Four bodies of masses 2, 3,5 and \(8 \mathrm{~kg}\) are placed at the four corners of a square of side \(2 \mathrm{~m}\). The position of \(\mathrm{CM}\) will be :
163791
Two bodies of masses \(\mathbf{5} \mathbf{~ k g}\) and \(\mathbf{2} \mathbf{~ k g}\) are moving with velocity \((\hat{i}+2 \hat{j}-2 \hat{k})\) and \((-2 \hat{i}-5 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) respectively. What is the velocity of the centre of mass:
163792
The position of centre of mass of a system consisting of two particles of masses \(m_1\) and \(m_2\) separated by a distance \(L\) apart, from \(m_1\) will be:
1 \(\frac{m_1 L}{m_1+m_2}\)
2 \(\frac{m_2 L}{m_1+m_2}\)
3 \(\frac{m_2}{m_1} L\)
4 \(\frac{L}{2}\)
Explanation:
it shows from the figure that \( X_{c m}=\frac{m_1 \times 0+m_2 L}{m_1+m_2}=\frac{m_2}{m_1+m_2} L \) \( Y_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) \( Z_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) centre of mass is at a distance \(\left\) from \(m_1\) internally on the line joining the two particles.
NCERT-XI-I -98
4 RBTS PAPER
163793
If the system is released, then acceleration of the center of mass of the system is :
1 \(g / 4\)
2 \(g / 2\)
3 \(\mathrm{g}\)
4 \(2 \mathrm{~g}\)
Explanation:
Acceleration of each mass \( a=\left(\frac{3 m-m}{3 m+m}\right) g=g / 2 \) \( a c c^n \text { of CM, } a_{c m}=\frac{m_1 a_1+m_2 a_2}{m_1+m_2} \) \( a_{c m}=\frac{3 m \frac{g}{2}-m \frac{g}{2}}{3 m+m}=\frac{g}{4} \)
NCERT-XI- I -99
4 RBTS PAPER
163794
Four bodies of masses 2, 3,5 and \(8 \mathrm{~kg}\) are placed at the four corners of a square of side \(2 \mathrm{~m}\). The position of \(\mathrm{CM}\) will be :
163791
Two bodies of masses \(\mathbf{5} \mathbf{~ k g}\) and \(\mathbf{2} \mathbf{~ k g}\) are moving with velocity \((\hat{i}+2 \hat{j}-2 \hat{k})\) and \((-2 \hat{i}-5 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) respectively. What is the velocity of the centre of mass:
163792
The position of centre of mass of a system consisting of two particles of masses \(m_1\) and \(m_2\) separated by a distance \(L\) apart, from \(m_1\) will be:
1 \(\frac{m_1 L}{m_1+m_2}\)
2 \(\frac{m_2 L}{m_1+m_2}\)
3 \(\frac{m_2}{m_1} L\)
4 \(\frac{L}{2}\)
Explanation:
it shows from the figure that \( X_{c m}=\frac{m_1 \times 0+m_2 L}{m_1+m_2}=\frac{m_2}{m_1+m_2} L \) \( Y_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) \( Z_{c m}=\frac{m_1 \times 0+m_2 \times 0}{m_1+m_2}=0 \) centre of mass is at a distance \(\left\) from \(m_1\) internally on the line joining the two particles.
NCERT-XI-I -98
4 RBTS PAPER
163793
If the system is released, then acceleration of the center of mass of the system is :
1 \(g / 4\)
2 \(g / 2\)
3 \(\mathrm{g}\)
4 \(2 \mathrm{~g}\)
Explanation:
Acceleration of each mass \( a=\left(\frac{3 m-m}{3 m+m}\right) g=g / 2 \) \( a c c^n \text { of CM, } a_{c m}=\frac{m_1 a_1+m_2 a_2}{m_1+m_2} \) \( a_{c m}=\frac{3 m \frac{g}{2}-m \frac{g}{2}}{3 m+m}=\frac{g}{4} \)
NCERT-XI- I -99
4 RBTS PAPER
163794
Four bodies of masses 2, 3,5 and \(8 \mathrm{~kg}\) are placed at the four corners of a square of side \(2 \mathrm{~m}\). The position of \(\mathrm{CM}\) will be :
