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
+
=
+

Where;
= mass of bullet = 7.00g = 0.007kg
= mass of tin can = 150g = 0.15kg
= initial velocity of bullet before impact = 200m/s
= initial velocity of tin can before impact = 0m/s (since the can is stationary)
= final velocity of the bullet after impact
= final velocity of the tin can after impact = 180cm/s = 1.8m/s
Substitute these values into the equation above;
=>
+
=
+

=> (0.007 x 200) + (0.15 x 0) = (0.007 x
) + (0.15 x 1.8)
=> 1.4 + 0 = 0.007
+ 0.27
=> 1.4 = 0.007
+ 0.27
=> 0.007
= 1.4 - 0.27
=> 0.007
= 1.13
Solve for
=>
= 1.13 / 0.007
=>
= 161.43m/s
Therefore, the speed of the bullet after impact (leaving the can) is 161.43m/s