2022-01-13

An 8.00 g bullet is fired into a 2.50 keg block that is initially at rest at the edge of a frictionless table of height 1.00 m (fig. P9.57). The bullet remains in the block, and after impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

nick1337

Expert2022-02-07Added 777 answers

The system's initial velocity in the y direction (block plus bullet) is zero. only the acceleration brought on by gravity.

Thus

$S=ut+\frac{1}{2}a{t}^{2}$

$1=0+\frac{1}{2}\times 9.81\times {t}^{2}$

$t=0.452$ sec

In y direction acceleration is zero.

$S=ut+\frac{1}{2}a{t}^{2}$

$2=u\times 0.452+\frac{1}{2}\times 0\times {t}^{2}$

$u=4.43$ m/s

This is initial velocity of system before falling down.

Now, apply conversation of momentum

${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=({m}_{1}+{m}_{2})u$

$(0.008)\times {v}_{1}+(0.250\times 0)=(0.258)\times 4.43$

${v}_{1}=\frac{0.258\times 4.43}{0.008}=142.86$ m/s

Thus, speed of bullet $=143$ m/s - Answer

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