1) A block of masss M is hanging vertically from a rope of length L. A bullet of

Question

1) A block of masss M is hanging vertically from a rope of length L. A bullet of mass m is shot at the blcok and embeds itself in the block. The acceleration of gravity is g. The speed of the bullet is unknown. If the block and bullet composite systme swing up to a final angle theta, compute an expression for the speed fo the bullet.
2) A block of mass M is attached to a spring of spring constant K. A bullet of mass m is shot at the block and embeds itself in the block. The speed of the bullet is unknown. If the block and bullet composite system compress the spring a distance b, compute an expression for the speed of the bullet. 1) A block of masss M is hanging vertically from a rope of length L. A bullet of mass m is shot at the blcok and embeds itself in the block. The acceleration of gravity is g. The speed of the bullet is unknown. If the block and bullet composite systme swing up to a final angle theta, compute an expression for the speed fo the bullet.
2) A block of mass M is attached to a spring of spring constant K. A bullet of mass m is shot at the block and embeds itself in the block. The speed of the bullet is unknown. If the block and bullet composite system compress the spring a distance b, compute an expression for the speed of the bullet.

Explanation / Answer

1)In this case we make use of law of conservation of linear momentum m vi1 +M vi2 = (m+ M) vf Here vi2 = 0, since initially the block is at rest    vi1 = (m+ M) vf/m--------(1) using conservation of energy , The potential energy at top of swing is equal to the kinetic energy at the bottom of the swing (m + M)gh = 1/2(m + M)vf^2 vf = v2gh substitute this value in eq(1) vi1 = (m + M) v2gh /m1 is the expression for speed of the bullet ------------------------------------------------------------------------------------------------------- 2)Using law of conservation of momentum    m vi1 + M vi2 = (m+ M) vf Here vi2 = 0, since initially the block is at rest vf = (m / m+M) vi1-------(1) using conservation of energy , the kinetic energy is equal to the elastic potential energy 1/2(m + M)vf^2 = 1/2kb^2---(2) substitute eq 1 in eq 2, we get vi1 = (b/m)vk(m+ M) is the speed of the bullet Here vi2 = 0, since initially the block is at rest vf = (m / m+M) vi1-------(1) using conservation of energy , the kinetic energy is equal to the elastic potential energy 1/2(m + M)vf^2 = 1/2kb^2---(2) substitute eq 1 in eq 2, we get vi1 = (b/m)vk(m+ M) is the speed of the bullet

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