An archer shoots a 2.1×10 2 kgarrow at a target with a speed of 57 m/s. When it

Question

An archer shoots a 2.1×102 kgarrow at a target with a speed of 57 m/s. When it hits the target,it penetrates to a depth of 9.0×102 m.

Part A: What was the magnitude of the average force exerted by thetarget on the arrow?
Part B: What was the direction of the average force exerted bythe target on the arrow? (Toward the direction of motion or opposite to the directionof motion?)

Part C: If the mass of the arrow is doubled, and the force exertedby the target on the arrow remains the same, by what multiplicativefactor does the penetration depth change?
(x2 / x1) = An archer shoots a 2.1×102 kgarrow at a target with a speed of 57 m/s. When it hits the target,it penetrates to a depth of 9.0×102 m.

Part A: What was the magnitude of the average force exerted by thetarget on the arrow?
Part B: What was the direction of the average force exerted bythe target on the arrow? (Toward the direction of motion or opposite to the directionof motion?)

Part C: If the mass of the arrow is doubled, and the force exertedby the target on the arrow remains the same, by what multiplicativefactor does the penetration depth change?
(x2 / x1) =

Explanation / Answer

W=Ke F S= 0.5mV2then F= 0.5mV2 /S F= 0.5x2.1x10-2(57.0)2 /9.0×102 F= 380N b) Opposite to the direction ofmotion since it the force that stopped the arrow.
c) we have F1= 0.5m1V2 /S1  and F2= 0.5m2V2 /S2  since F1= F2  and   2m1= m2   we have   0.5m1V2 /S1  = 0.5m2V2 /S2      m1 /S1  = 2m1 /S2       S2/ S1 = 2m1/  m1
S2/S1 = 2

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