A 10 g bullet traveling at 450 m/s strikes a 10 kg , 1.1-m-wide door at the edge

Question

A 10 g bullet traveling at 450 m/s strikes a 10 kg , 1.1-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

Part A

What is the angular velocity of the door just after impact?

Express your answer to two significant figures and include the appropriate units.

Part A

What is the magnitude of the angular momentum of the 3.10 kg , 5.20-cm-diameter rotating disk in the figure (Figure 1) ?

The three masses shown in the figure(Figure 1) are connected by massless, rigid rods. Assume that m1 = 150 g and m2 = 300 g .

Part A

What is the x-coordinate of the center of mass?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the y-coordinate of the center of mass?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

angular velocity of door just after impact = velocity of door after impact/ distance of edge from hinge = v/r

we need to find velocity of door after impact first

we can find it using conservation of momentum

initial momentum of bullet = final momentum of bullet + door

(10 x 10^-3)(450) = (10 x 10^-3 +10)v

v = 0.4496 m/s

angular velocity of door after impact is = v/r = 0.4496/1.1 = 0.4087 rad/s

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