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In a ballistics test, a 28-g bullet pierces a sand bag that is 30cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag moving at 18m/s, what was the magnitude of the friction force (assuming it to be constant and the only force present) that the bullet experienced while it traveled through the bag

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Final answer:

The magnitude of the friction force experienced by the bullet can be calculated using the principle of conservation of momentum.

Step-by-step explanation:

The magnitude of the friction force experienced by the bullet can be calculated using the principle of conservation of momentum.

Since the only force acting on the bullet is the friction force, it can be equated to the change in momentum of the bullet:

Friction force = (final momentum - initial momentum) / time

Given that the initial velocity of the bullet is 55 m/s and the final velocity is 18 m/s, the initial momentum is 28 g * 55 m/s and the final momentum is 28 g * 18 m/s. The time can be calculated using the distance traveled and the average velocity which is the sum of the initial and final velocities divided by 2.

Using these values, the magnitude of the friction force can be calculated.

User Claireablani
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3 votes

Answer:

Step-by-step explanation:

Given that,

The mass of the bullet is

M= 28g = 0.028kg.

Thickness oof sack bag

d = 30cm = 0.3m

Initial velocity of bullet Vi = 55m/s

Velocity at which the bullet emerges the sack bag is Vf = 18m/s

We want to calculate the frictional force present in the sack bag.

Using conservation of energy.

Work done by friction in the sack is equal to change in kinetic energy.

Work done by friction is given as

Wf = Fr × d

The distance is the thickness of the bag

Wf = Fr × 0.3

Wf = 0.3 • Fr

Change in kinetic energy is calculated by

∆K.E = ½mVf² — ½ mVi²

∆K.E = ½m(Vf² — Vi²)

∆K.E = ½ × 0.028 ( 18²-55²)

∆K.E = 0.014 × -2701

∆K.E = -37.814 J

Since

Work done by friction = ∆K.E

0.3 Fr = -37.814

Fr = -37.814/0.3

Fr = -126.05 N

The, negative sign show that frictional force is opposing the motion.

Fr = 126.05 N

User Deepthi
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