A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1m/s so as to have same K.E. as that of the boy. The original speed of the man will be 

(1) $\sqrt{2}m/s$

(2) $(\sqrt{2}-1)m/s$

(3) $\frac{1}{(\sqrt{2}-1)}m/s$

(4) $\frac{1}{\sqrt{2}}m/s$ 

A particle of mass m at rest is acted upon by a force F for a time t. Its Kinetic energy after an interval t is 

(1) $\frac{F^2{t}^2}{m}$

(2) $\frac{F^2{t}^2}{2m}$

(3) $\frac{F^2{t}^2}{3m}$

(4) $\frac{Ft}{2m}$

Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear momentums is

(1) 4 : 1

(2) 1 : 1

(3) 1 : 2

(4) 1 : 4

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 

(1) m₁/m₂

(2) 1

(3) m₁ v₂/m₂ v₁

(4) m₂/m₁

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) 1.07 kJ

(2) 2.14 kJ

(3) 2.4 kJ

(4) 4.8 kJ

A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum but their different kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then 

(1) E₁ < E₂

(2) $\frac{E_1}{E_2}=\frac{m_1}{m_2}$

(3) E₁ > E₂

(4) E₁ = E₂

Four particles are given, having same momentum. Which one has maximum kinetic energy- 

(1) Proton

(2) Electron

(3) Deutron

(4) α-particles

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 

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

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

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

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

A block of mass m initially at rest is dropped from a height h on to a spring of force constant k. The maximum compression in the spring is x then-

(1) $mgh=\frac{1}{2}k{x}^2$

(2) $mg(h+x)=\frac{1}{2}k{x}^2$

(3) $mgh=\frac{1}{2}k{(x+h)}^2$

(4) $mg(h+x)=\frac{1}{2}k{(x+h)}^2$

A spherical ball of mass \(20\) kg is stationary at the top of a hill of height\(100\) m. It slides down a smooth surface to the ground, then climbs up another hill of height \(30\) m and finally slides down to a horizontal base at a height of \(20\) m above the ground. The velocity attained by the ball is: 
1. \(10 \) m/s
2. \(10 \sqrt{30} \) m/s
3. \(40 \) m/s
4. \(20 \) m/s