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

User Dinsen
by
7.0k points

2 Answers

4 votes

Answer:

160 m/s

Step-by-step explanation:

Momentum is conserved:

mu = mv + MV

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

v = 160 m/s

User Shebaw
by
6.7k points
5 votes

Answer:

161.43m/s

Step-by-step explanation:

Using the principle of conservation of linear momentum i.e

Total momentum before impact is equal to total momentum after impact.

=> Momentum of bullet before impact + Momentum of tin before impact

=

Momentum of bullet after impact + Momentum of tin after impact

i.e


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

Substitute these values into the equation above;

=>
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

Solve for
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

User Feralamillo
by
7.4k points