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2 years ago

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A 7.00 g bullet moving horizontally at 200 m/s strikes and passes through a 150 g tin can sitting on a post. Just after the impa

ct, the can has a horizontal speed of 180 cm/s. What was the bullet’s speed after leaving the can?

2 answers:

Orlov [11]2 years ago

8 0

<h2>Answer:</h2>

161.43m/s

<h2>Explanation:</h2>

Using the principle of conservation of linear momentum i.e

<em>Total momentum before impact is equal to total momentum after impact.</em>

<em>=> Momentum of bullet before impact + Momentum of tin before impact</em>

<em> = </em>

<em> Momentum of bullet after impact + Momentum of tin after impact</em>

<em> i.e </em>

$m_{B}$ $u_{B}$ + $m_{T}$ $u_{T}$ = $m_{B}$ $v_{B}$ + $m_{T}$ $v_{T}$

Where;

$m_{B}$ = mass of bullet = 7.00g = 0.007kg

$m_{T}$ = mass of tin can = 150g = 0.15kg

$u_{B}$ = initial velocity of bullet before impact = 200m/s

$u_{T}$ = initial velocity of tin can before impact = 0m/s (since the can is stationary)

$v_{B}$ = final velocity of the bullet after impact

$v_{T}$ = final velocity of the tin can after impact = 180cm/s = 1.8m/s

<em>Substitute these values into the equation above;</em>

=> $m_{B}$ $u_{B}$ + $m_{T}$ $u_{T}$ = $m_{B}$ $v_{B}$ + $m_{T}$ $v_{T}$

=> (0.007 x 200) + (0.15 x 0) = (0.007 x $v_{B}$ ) + (0.15 x 1.8)

=> 1.4 + 0 = 0.007 $v_{B}$ + 0.27

=> 1.4 = 0.007 $v_{B}$ + 0.27

=> 0.007 $v_{B}$ = 1.4 - 0.27

=> 0.007 $v_{B}$ = 1.13

<em>Solve for </em> $v_{B}$

=> $v_{B}$ = 1.13 / 0.007

=> $v_{B}$ = 161.43m/s

Therefore, the speed of the bullet after impact (leaving the can) is 161.43m/s

Vikki [24]2 years ago

3 0

Answer:

160 m/s

Explanation:

Momentum is conserved:

mu = mv + MV

(7.00) (200) = (7.00)v + (150) (1.8)

v = 160 m/s

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