A bullet of mass 4 g moving with an initial speed 300 m/s is fired into and pass

Question

A bullet of mass 4 g moving with an initial speed 300 m/s is fired into and passes through a block of mass 9 kg, as shown in the figure. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 943 N/m. If the block moves a distance 0.38 cm to the right after the bullet passed through it, find the speed v at which the bullet emerges from the block. Answer in units of m/s. Find the magnitude of the energy lost in the collision. Answer in units of J.

Explanation / Answer

Initial KE of the bullet, KEbi = 0.5*0.004*300^2 = 180 J

final KE of bullet, KEbf = 0.5*0.004*v^2

Energy stored in spring, PE = 0.5*943*0.0038^2 = 0.0068 J

Let the speed of the 9kg mass just after collision be v'

So, By conservation of energy , KE of mass = PE in spring

So, 0.5*M*v'^2 = 0.5*k*x^2 = 0.0068

So, M*v' = sqrt(k*x^2*M)

Now, by conservation of momtntum,

m*u = m*v+ M*v'

where u = initial speed of bullet = 300 m/s

v = speed of bullet after collision

So, 0.004*300 = 0.004*v + sqrt(943*0.0038^2*9)

So, v = 212.4 m/s <-----answer

So, energy lost = 0.5*0.004*(300^2 - 212.4^2) - 0.5*943*0.0038^2

= 89.8 J <-----answer

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