Question

A 3.50-g bullet has a muzzle velocity of 250 m/s when fired by a rifle with a weight of 25.0 N. (a) Determine the recoil speed of the rifle. m/s (b) If a marksman with a weight of 650 N holds the rifle firmly against his shoulder, determine the recoil speed of the shooter and rifle.

Answer 1

Answer:

(a) 0.343 m/s

(b) 0.012 m/s

Explanation:

(a) From the question above,

MV = mv............................... Equation 1

Where M = mass of the rifle, V = recoiling speed of the rifle, m = mass of the bullet, v = velocity of the bullet.

make V the subject of the equation

V = mv/M........................... Equation 2

Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg.

Substitute into equation 2

V = (0.0035×250)/2.55

V = 0.343 m/s.

(b) Similarly,

(M'+M)V' = mv....................... Equation 3

Where M' = mass of the marksman, V' = recoiling speed of the shooter and rifle

make V' the subject of the equation

V' = mv/(M'+M)................... Equation 4

Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg, M' = 650 N = 650/9.8 = 66.33 N

Substitute into equation 4

V' = (0.0035×250)/(66.33+2.55)

V' = 0.8125/68.88

V' = 0.012 m/s

Answer 2

The recoil velocity can be obtained using the principle of conservation of linear momentum.

Using the principle of conservation of linear momentum;

momentum before collision = momentum after collision

Mass of the bullet = 3.50-g or 0.0035 Kg

Mass of the rifle = 2.5 Kg

Where;

M1 = mass of rifle

M2 = mass of bullet

u1 =initial velocity of rife

u2 = initial velocity of the bullet

(2.5 × 0) + (0.0035 × 250) = (0.0035 × 0) + (2.5 × v)

0.875 = 2.5 v

v = 0.35 m/s

For the shooter and the rifle;

(67.5 × 0) + (0.0035 × 250) =  (0.0035 × 0) + (67.5 × v)

0.875 = 67.5 × v

v = 0.013 m/s

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