Question

A 44g bullet strikes and becomes embedded in a 1.54 kg block of wood placed on a horizontal surface just in front of the gun . If the coefficient of kinetic friction between the block and the surface is 0.28 , and the impact drives the block a distance of 18.Om before it comes to rest , what was the muzzle speed of the bullet ?

Answer 1

Answer:

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Le v1 be the velocity of the bullet of mass m1 and v2 be the velocity of the block of mass m2 after hit.

Le v1 be the velocity of the bullet of mass m1 and v2 be the velocity of the block of mass m2 after hit.Therefore, the kinetic energy of of the block and bullet together =(1/2)(m1+m2)v2^2 which is used up in pushing the block with aforce of (m1+m2)g* coefficient of Kinetic friction, M* times the displacement, s of the block after hit by bullet = (m1+m2)g*M*s. Therefore,

Le v1 be the velocity of the bullet of mass m1 and v2 be the velocity of the block of mass m2 after hit.Therefore, the kinetic energy of of the block and bullet together =(1/2)(m1+m2)v2^2 which is used up in pushing the block with aforce of (m1+m2)g* coefficient of Kinetic friction, M* times the displacement, s of the block after hit by bullet = (m1+m2)g*M*s. Therefore,(1/2)(m1+m2) v2^2 = (m1+m2)g*M*s

Le v1 be the velocity of the bullet of mass m1 and v2 be the velocity of the block of mass m2 after hit.Therefore, the kinetic energy of of the block and bullet together =(1/2)(m1+m2)v2^2 which is used up in pushing the block with aforce of (m1+m2)g* coefficient of Kinetic friction, M* times the displacement, s of the block after hit by bullet = (m1+m2)g*M*s. Therefore,(1/2)(m1+m2) v2^2 = (m1+m2)g*M*sSo, v2^2 = 2g×M×s .......................(1)

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/s

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.Since the momentums are conserved,

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.Since the momentums are conserved,m1v1 = (m1+m2)v2 or

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.Since the momentums are conserved,m1v1 = (m1+m2)v2 orv1 = (m1+m2)v2 / m1 , substituting m1= 15 g or 15/1000 kg = 0.015 kg, m2 = 1.24 kg, v2 = sqrt(60.4296) , we get,

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.Since the momentums are conserved,m1v1 = (m1+m2)v2 orv1 = (m1+m2)v2 / m1 , substituting m1= 15 g or 15/1000 kg = 0.015 kg, m2 = 1.24 kg, v2 = sqrt(60.4296) , we get,v1=(0.015+1.24)sqrt(60.4296)/(0.015)

s .......................(1)So, v2 = sqrt(2*9.81*0.28*11) = sqrt(60.4296) = 7.7736 m/sTherefore, acceleration a = (0-V2^2)/(2s) = -2g*M*s/(2s)=-g*M=-9.81*0.28 = -0.27468m/s^2 is the acceleration or 0.27468m/s^2 is the retardation or deceleration of the block.Since the momentums are conserved,m1v1 = (m1+m2)v2 orv1 = (m1+m2)v2 / m1 , substituting m1= 15 g or 15/1000 kg = 0.015 kg, m2 = 1.24 kg, v2 = sqrt(60.4296) , we get,v1=(0.015+1.24)sqrt(60.4296)/(0.015)= 650.3952 m/s is the speed of the bullet.

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