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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
If E are const. then = 2
If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is
⇒
⇒ of E
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
⇒
⇒
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
Both fragment will possess the equal linear momentum
⇒ ⇒
∴ Total energy of system
=
= 4800 J = 4.8 kJ
An object of mass 3m splits into three equal fragments. Two fragments have velocities and . The velocity of the third fragment is
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 vj and vi, the third fragment must have velocity -v(i + j) to conserve momentum.
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
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.
A shell initially at rest explodes into two pieces of equal mass, then the two pieces will
According to law of conservation of linear momentum both pieces should possess equal momentum after explosion. As their masses are equal therefore they will possess equal speed in opposite direction.
Two solid rubber balls A and B having masses 200 gm and 400 gm respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest, then the velocity of B is
In a completely inelastic collision between two bodies, the total momentum before collision is equal to the total momentum after collision. Given: m1 = 200 g = 0.2 kg, v1 = 0.3 m/s, m2 = 400 g = 0.4 kg, v2 = ? Using conservation of momentum: m1v1 + m2v2 = (m1 + m2)v, where v is the final common velocity. Substituting the values, we get 0.20.3 + 0.4v = 0.6*0, which gives v = -0.15 m/s.
A cannon ball is fired with a velocity 200 m/sec at an angle of 60° with the horizontal. At the highest point of its flight ,it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity
A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.