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‘The linear momentum is the product of mass and velocity. The, momentum is a vector and its direction is same as that of the, velocity., , p=mi, , |p| =p=mv, Relation between Momentum and Kinetic Energy, , 1 ig, my, K= > my, pre, 2, , 2, , K=2, 2m, , p= 2mk, Principle of Conservation of Linear Momentum, , In the absence of any external force or when the external forces, can be neglected in comparison to the internal forces, the linear, momentum of system of particles remains the same., , , , , , Since for (gun + bullet) the system, the external force is zero., Hence by momentum conservation, , Initial momentum = Final momentum, , O=mu+Mv, Mass of bullet, m = 50 g = 50x 10 kg, u = 200 mis, Mass of gun, M=2kg, , 0=50x 10° x 200+2v, , v=-5 m/s, where the —ve sign indicates that the velocity of gun is, opposite to that of the bullet.
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By momentum conservation,, O=mv+3m/ = v= 5, , Given K; = Fm? = K, Ky= 5x3? =, , x ey i, £ xsi =9(3) 3, , R=, , OR, 0, , a* i Q-« a, 4m m 3m, , Due to the momentum conservation, momentum of two, pieces is equal and opposite,, , | ee, fg 2, oa om, Fi oo., ro "sg, , , , Solution:, When the smaller block is moving on the vertical part, the, (m + M) system will move with velocity vg in horizontal, direction., , +—+%, , , , Since no external force in horizontal direction, hence by the, momentum conservation, , my = (m+ M)vy, , % mv 50x20 1, , ~m+M (50+1950) 2, , , , , , 2m vcos 6 x, , At the highest point, the velocity is in horizontal direction, and equal to v cos @. One piece retraces its path, i.e. its, velocity is v cos @ in opposite direction as shown. Let the, velocity of other piece is v’,, Applying the momentum conservation at the higher point, 2mv cos 8=~mv cos 6+ mv’, v’=3vcos @, , Solution:, , Mass number = sum of the number of protons and neutrons, A=Z+N, , Z: number of protons, , N: number of neutrons, , Let m, = my =m, , U**; mass; 238m, , Th™*; mass: 234m, , particle (;He"): mass: 4m, , 5 © Isle, , FO", F Ort
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Solution:, ‘Total mass 5 m is split in a ratio 1 ; | : 3, hence masses of, three pieces are m, =m, m, =m and m, = 3m., sin @, y - ; “ ,, L. ©C) . l 8 v0, x sm m 3m, Let the velocity of heavier piece is v at an angle 6. Here,, we have to conserve momentum in the x- and y-direction, separately., By momentum conservation, x-direction: Sm x 0 = mvg + 3my cos 6, v, @= 2, ¥COs 3, y-direction: Sm x 0 = mv + 3mv sin 8, , vsin @= —“0, , The —ve sign indicates that the velocity is in opposite, direction, , 3m, , , , %, , 135° %, , a, s, , ‘The direction of the third piece is at an angle 135° from, , either of smaller pieces., Initial K-E. = 0, 2, 1 1 1 vv, Final KE, = ra +m fan[ 2), , K.E. released in explosion = ; mya, , OR, This problem can be solved without conserving momentum, conservation in x- and y-direction separately., Since the bomb is at rest, hence initial momentum is zero, and hence the final momentum of system of pieces will be, zero, due to the momentum conservation., , ‘The resultant momentum of smaller pieces will be equal and, Opposite to the momentum of heavier piece., , 0 Yo, ©... 4 ow, 5m nom, , Let m vo =p, , pg, Lae, p, , Resultant momentum P = vip, , 6=45°, Hence the momentum of heavier piece is equal and opposite, toR, , P, , , , P= Vip = 2mv,, , P=3mv, , 3my = J2mvy, fin, , 3, , v: Velocity of heavier piece.
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Solution:, Initial momentum of a system of particle is zero,, my, 120° mv, , , , Velocity of the shell w.r.t. the ground = (vp cos a@—v), , v: velocity of cannon w.x.t. the ground, 420°\}120° Applying momentum conservation in the horizontal direction, ‘ ae 0 =m,(v9 cos a— v) — myv, After the collision, the momentum of system of particles @, should be zero. The momentum of A becomes zero after the career *, collision. The final momentum of system should be zero, for es, , this C should move opposite to B with the same speed v., , , , 6 kg 2kg akg, , , , , , A 8, By the momentum conservation, 0=2v, —4y,, itasonangs ¥ = 2vy @, Yp: Vayp: Velocity of boy relative to platform Given relative velocity, V: ¥pyg: velocity of platform relative to ground vy +¥,=30 (ii), Vang =%-¥ Solving, v, = 10 m/s, v, = 20 m/s, By the momentum conservation o 4 5, 0=m(vp— v)- Mv 4 >, a iy, bi b—— dg ——_ +t —— 6, ———1, Retardation on the rough surface, , a= ENE = yg =05x 10=5 m/s?, , , , For 2 kg: v=? ~2as
