An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal

Question

An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 4.8 cm. The force constant of the spring is 1800 N/m. In the figure, the impulse of the block (including the bullet), due to the spring, during the entire time interval in which block and spring are in contact is closest to:

An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 4.8 cm. The force constant of the spring is 1800 N/m. In the figure, the impulse of the block (including the bullet), due to the spring, during the entire time interval in which block and spring are in contact is closest to:

Explanation / Answer

So
86.4 = 4.008 a
a = 86.4 / 4.008 = 21.55m/s^2

Now we can use this information to work out the velocity of the bullet/block combination.
From the equations of motion we have

v^2 = u^2 + 2 ax
where v = final velocity (m/s); u = initial velocity (m/s); a = acceleration (m/s^2); x = distance (m)

Since the surface is frictionless, no energy is lost so once the string is back to its natural length, the bullet/block combination will have the same velocity it had when it struck the spring.

v^2 = 0 + (2 *21.55 * 0.048)
v = 1.438 m/s

We can now go back to the momentum equation to work out the velcoity of the bullet

0.008 v1 = (4.008) * 1.438
v1 = 720.438 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.