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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
A 238U nucleus decays by emitting an alpha particle of speed v ms–1. The recoil speed of the residual nucleus is (in ms–1)
A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest. After collision their final velocities are V and v respectively. The value of v is
A shell of mass m moving with velocity v suddenly breaks into 2 pieces. The part having mass m/4 remains stationary. The velocity of the other shell will be
Two equal masses m1 and m2 moving along the same straight line with velocities +3 m/s and –5 m/s respectively collide elastically. Their velocities after the collision will be respectively
In an elastic collision between two bodies, the total momentum before and after the collision remains the same, and the total kinetic energy is also conserved. Using these principles and the given initial velocities, we can calculate the final velocities to be -5 m/s and +3 m/s.
A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m. Then the body rebounds to a height of
=
A body at rest breaks up into 3 parts. If 2 parts having equal masses fly off perpendicularly each after with a velocity of 12m/s, then the velocity of the third part which has 3 times mass of each part is
Using the conservation of momentum principle, we can equate the initial momentum (which is zero) to the final momentum, which is the vector sum of the momenta of the three parts. This gives the velocity of the third part as 4√2 m/s, making an angle of 135° with each of the other two parts.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.