Thanks You may use the provided equation sheet and your calculator. You MUST sho

Question

Thanks

You may use the provided equation sheet and your calculator. You MUST show work to receive full credit on the first three problems. Work problems in the space provided. If more space is required you may work on the back side attach additional sheets. Answer multiple choice questions on the scantron form. Don't forget your name and on the subject your test version. Multiple choice questions are worth 2.5 points each, all other point values are as indicated. Please read each question carefully and good luck! Consider the ballistic pendulum shown in the figure below. A bullet of mass m = 9.1 g is traveling with of vb immediately before coming into contact with a block of mass M = 5.2 kg. After the bullet is lodge block the bullet-block system swings upward to a max height of h = 6.5 cm Before the collision After the collision During the bullet-block collision, is the total linear momentum of the bullet-block system served During the bullet-block collision, is the total kinetic energy of the bullet-block system conserve What is the speed of the bullet-block system (Vm+ M) after the collision?

Explanation / Answer

a)

yes the total linear momentum is conserved since the collision is inelastic and in inelastic collision , momentum remain conserved

b)

in Inelastic collisions , the kinetic energy is not conserved as some energy is lost

c)

h = height gained = 6.5 cm = 0.065 m

Vm + M = speed of combination just after the collision

Using conservation of energy

Kinetic energy = Potential energy

(0.5) (M + m) V2m + M = (M + m) gh

Vm + M = sqrt(2 gh) = sqrt(2 x 9.8 x 0.065) = 1.13 m/s

d)

using conservation of momentum

m Vb + M (0) = (M+ m) Vm + M

(0.0091)Vb = (5.2 + 0.0091) (1.13)

Vb = 646.84 m/s

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