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Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of their linear momenta is then

3 : 1

9 : 1

1 : 1

$\sqrt{3} : 1$

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Explanation

$P = \sqrt{2 m E}$ ∴ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$ (if E = constant)

∴ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{3}{1}}$

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Two bodies with kinetic energies in the ratio of 4 : 1 are moving with equal linear momentum. The ratio of their masses is

1 : 2

1 : 1

4 : 1

1 : 4

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Explanation

$E = \frac{p^{2}}{2 m}$ ∴ $m \propto \frac{1}{E}$ (If momentum are constant)

$\frac{m_{1}}{m_{2}} = \frac{E_{2}}{E_{1}} = \frac{1}{4}$

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If the kinetic energy of a body becomes four times of its initial value, then new momentum will

Becomes twice its initial value

Become three times its initial value

Become four times its initial value

Remains constant

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Explanation

$P = \sqrt{2 m E}$

∴ $P \propto \sqrt{E}$ i.e. if kinetic energy becomes four time then new momentum will become twice.

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A bullet is fired from a rifle. If the rifle recoils freely, then the kinetic energy of the rifle is

Less than that of the bullet

More than that of the bullet

Same as that of the bullet

Equal or less than that of the bullet

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Explanation

$E = \frac{P^{2}}{2 m} .$ If P = constant then $E \propto \frac{1}{m}$

i.e. kinetic energy of heavier body will be less. As the mass of gun is more than bullet therefore it possess less kinetic energy.

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A 4 kg mass and a 1 kg mass are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

1 : 2

1 : 1

2 : 1

4 : 1

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Explanation

$P = \sqrt{2 m E} .$ If E are const. then $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{4}{1}}$ = 2

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If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is

150%

200%

225%

300%

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Explanation

$E = \frac{P^{2}}{2 m}$ ⇒ $E_{2} = E_{1} {(\frac{P_{2}}{P_{1}})}^{2} = E_{1} {(\frac{2 P}{P})}^{2}$

⇒ $E_{2} = 4 E = E + 3 E = E + 300 %$ of E

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A stationary particle explodes into two particles of masses m₁ and m₂ which move in opposite directions with velocities v₁ and v₂. The ratio of their kinetic energies E₁/E₂ is

m₁/m₂

m₁ v₂/m₂ v₁

m₂/m₁

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Explanation

$E = \frac{P^{2}}{2 m}$

⇒ $E \propto \frac{1}{m}$

⇒ $\frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}}$

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A bomb of mass 3.0 Kg explodes in air into two pieces of masses 2.0 kg and 1.0 kg. The smaller mass goes at a speed of 80 m/s.The total energy imparted to the two fragments is

1.07 kJ

2.14 kJ

2.4 kJ

4.8 kJ

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Explanation

Both fragment will possess the equal linear momentum

$m_{1} v_{1} = m_{2} v_{2}$ ⇒ $1 \times 80 = 2 \times v_{2}$ ⇒ $v_{2} = 40 m . / s$

∴ Total energy of system $= \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

= $\frac{1}{2} \times 1 \times {(80)}^{2} + \frac{1}{2} \times 2 \times {(40)}^{2}$

= 4800 J = 4.8 kJ

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An object of mass 3m splits into three equal fragments. Two fragments have velocities $v \hat{j}$ and $v \hat{i}$ . The velocity of the third fragment is

$v (\hat{j} - \hat{i})$

$v (\hat{i} - \hat{j})$

$- v (\hat{i} + \hat{j})$

$\frac{v (\hat{i} + \hat{j})}{\sqrt{2}}$

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Explanation

This is a problem of conservation of momentum. The total initial momentum of the 3m mass is zero. After splitting, the sum of momenta of the three fragments must also be zero. If two fragments have velocities v_j and v_i, the third fragment must have velocity -v(i + j) to conserve momentum.

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A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms^–1. The kinetic energy of the other mass is

256 J

486 J

524 J

324 J

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Explanation

According to the principle of conservation of momentum, the total momentum before the explosion must be equal to the total momentum after the explosion. Given: m1 = 18 kg, v1 = 6 m/s, m2 = 12 kg. Using p1 + p2 = 0, we get m2v2 = -m1v1 = -18*6 = -108 kg.m/s. Therefore, v2 = -108/12 = -9 m/s. The kinetic energy of the 12 kg mass is (1/2)mv^2 = (1/2)12(-9)^2 = 486 J.

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