# A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet.?

If the bullet has a mass of 27.8 g and a speed of 230 m/s , how high will the block rise into the air after the bullet becomes embedded in it?

PLEASE HELP! i used the equation for conservation of momentum and I keep getting 106m...theres no way thats right! 🙁

• If the bullet has a mass of 27.8 g and a speed of 230 m/s , how high will the block rise into the air after the bullet becomes embedded in it? gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet.?

Let’s use the following equation to determine the initial momentum.

M = 0.0278 * 230 = 6.394

Total mass = 0.0278 + 1.4 = 1.4278 kg

Final momentum = 1.4278 * v

1.4278 * v = 6.394

v = 6.394 ÷ 1.4278

This is approximately 4.478 m/s. As the block and bullet rise to their maximum height, their velocity deceases from this number to 0 m/s at the rate of 9.8 m/s each second. Let’s use the following equation to determine their maximum height.

vf^2= vi^ 2 + 2 * a * d, vf = 0, a = -9.8

0 = (6.394 ÷ 1.4278)^2 + 2 * -9.8 * d

19.6 * d = (6.394 ÷ 1.4278)^2

d = (6.394 ÷ 1.4278)^2 ÷ 19.6

This is approximately 1.023 meters.

• M1V1 = -M2V2, to get the velocity the bullet in the wood will go

V2= M1V1 (bullets momentum) / (M1+M2) (the masses stuck together)

V2= (0.0278 x 230) / (1.4+0.0278), = 4.48 m/s

Now there are many formulas we can use to get different values, but now that the wood and bullet are going up against gravity at that speed, the distance can be found by:

(where r = displacement, u = initial velocity and the velocity at peak height, v, = 0 )

V^2 - U^2 = 2ar

1/2 x (0 - 4.48^2 / -9.8 ) = r

r = 1.024 metres

*As a bonus:*

To check if your answer is right, you can use the formula

mu1 + mu2 = mv1 + mv2

and rearrange that formula for values you got at the start, for example, mu2 = 0, so to check it:

mu2, 0, = mv1 + mv2 - mu1

= 0.002

So it is right. it is only not 0.0 due to small rounding off errors

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