A 0.025 kg bullet is fired into a 10 kg block initially at rest on the horizonta

Question

A 0.025 kg bullet is fired into a 10 kg block initially at rest
on the horizontal portion of a frictionless surface as shown. The
speed of the bullet as it was about to enter the block was 802
m/s. Find:
[a] the speed of the bullet and block just after the bullet entered
the block
[b] the height y to which the center of mass of the system rises
[c] the energy lost by the system during the collision of the
bullet and the block

A 0.025 kg bullet is fired into a 10 kg block initially at rest on the horizontal portion of a frictionless surface as shown. The speed of the bullet as it was about to enter the block was 802 m/s. Find: the speed of the bullet and block just after the bullet entered the block the height y to which the center of mass of the system rises the energy lost by the system during the collision of the bullet and the block

Explanation / Answer

m1 = 0.025 kg, v1 = 802 m/s

m2 = 10 kg



a) let v is the speed ogf the system after the bbullet embeded

m1*v1 = (m1+m2)*v


v = (m1*v1)/(m1+m2)

= 0.025*802/(0.025+10)

= 2 m/s

b)

h = v^2/2*g

= 2^2/(2*9.8)

= 0.204 m

c) Ki = 0.5*m1*v1^2 = 0.5*0.025*802^2 = 8040.05

Kf = 0.5*m1+m2)v^2 = 0.5*(0.025+10)*2^2 = 20.05 J

ki-kf = 8020 J

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