Question 1:
easy
A 1 kg mass has a kinetic energy of 1 joule when its speed is :
Question 2:
easy
When the velocity of a body is doubled :
Momentum of an object is P = m.v
So, when velocity is doubled momentum also doubles
Question 3:
easy
If the kinetic energy of a body increases by 800%, its momentum increases by
The momentum \(P\) is related to kinetic energy \(K\) by \(P = \sqrt{2mK}\). An 800% increase means the new kinetic energy is \(K' = 9K\). Thus, the new momentum is \(P' = \sqrt{9} P = 3P\), representing an increase of 200%.
Question 4:
easy
Two masses \( 4m \) and \( 9m \) move with equal kinetic energy. The ratio of the magnitude of their momenta is:
Since kinetic energy \( K \) is the same for both masses, the momentum is proportional to the square root of the mass, \( p = \sqrt{2mK} \). Thus, the ratio of their momenta is \( \frac{p_1}{p_2} = \sqrt{\frac{4m}{9m}} = \frac{2}{3} \).
Question 5:
easy
Two masses \(1\text{ g}\) and \(9\text{ g}\) are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is
Since linear momentum \(p = \sqrt{2mK}\) and kinetic energy \(K\) is constant, \(p \propto \sqrt{m}\). Therefore, the ratio of momenta is \(\frac{p_1}{p_2} = \sqrt{\frac{1}{9}} = 1:3\).
Question 6:
easy
Assertion (A): A body cannot have kinetic energy without having linear momentum but it can have momentum without having mechanical energy.
Reason (R): Linear momentum and energy have same dimensions.
Assertion (A) is false:
If a body has linear momentum (\(p \neq 0\)), it must have velocity (\(v \neq 0\)), which implies it must also have kinetic energy (\(KE = \frac{1}{2}mv^2 \neq 0\)). Since kinetic energy is a component of mechanical energy, it cannot have momentum without mechanical energy.
Reason (R) is false: Linear momentum has dimensions \(MLT^{-1}\) while energy has dimensions \(ML^2T^{-2}\), which are different.
Question 7:
easy
Assertion (A): Kinetic energy of a system can be increased without applying any external force on the system.
Reason (R): If external forces are absent then work done by internal forces is equal to change in kinetic energy.
According to the work-energy theorem, `\(W_{net} = \Delta KE\)`. If external forces are absent, the net work done on the system is only due to internal forces, i.e., `\(W_{int} = \Delta KE\)`. Thus, internal forces can increase kinetic energy, for example, in an explosion. Both assertion and reason are true, and the reason correctly explains the assertion.
Question 8:
easy
Arun has \(\left(\frac{1}{3}\right)^{\text{rd}}\) of the kinetic energy as that of Raunak when they both run. If Raunak has half the mass of Arun, then the relationship between the speed of Arun \(v’\) and speed of Raunak \(v\) is
Let \(m_A = 2m_R\). Given \(K_A = \frac{1}{3}K_R â \frac{1}{2}m_A v'^2 = \frac{1}{3}\left(\frac{1}{2}m_R v^2\right)\). Substituting \(m_A = 2m_R\), we get \(2v'^2 = \frac{1}{3}v^2 â v' = \frac{v}{\sqrt{6}}\).
Question 9:
easy
A particle of mass \(0.4\text{ kg}\) is moving with a velocity of \((6\hat{i} – 8\hat{j})\text{ m/s}\), then the kinetic energy of the particle is
The speed squared is \(v^2 = 6^2 + (-8)^2 = 100\text{ m}^2\text{/s}^2\). The kinetic energy is \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.4)(100) = 20\text{ J}\).
Question 10:
easy
A bomb of mass \(30\text{ kg}\) at rest explodes into two pieces of masses \(18\text{ kg}\) and \(12\text{ kg}\). The velocity of \(18\text{ kg}\) mass is \(6\text{ ms}^{-1}\). The kinetic energy of the other mass is:
(2005)
By conservation of momentum, \(m_1v_1 + m_2v_2 = 0\) (since initial momentum is zero). Given \(m_1 = 18\text{ kg}\), \(v_1 = 6\text{ m/s}\), and \(m_2 = 12\text{ kg}\). So, \(18 \times 6 + 12v_2 = 0 \Rightarrow 108 + 12v_2 = 0 \Rightarrow v_2 = -9\text{ m/s}\). The kinetic energy of the other mass is \(KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(12)(-9)^2 = 6 \times 81 = 486\text{ J}\).