# Chapter 7 - Problems - Page 266: 57

The block lands on the floor a distance of $0.494~m$ from the table.

#### Work Step by Step

By conservation of momentum, the final momentum of the block and bullet is equal to the initial momentum of the bullet. We can find the speed $v_f$ of the block and bullet just after the collision: $m_f~v_f = m_0~v_0$ $v_f = \frac{m_0~v_0}{m_f}$ $v_f = \frac{(0.010~kg)(400.0~m/s)}{0.010~kg+4.0~kg}$ $v_f = 0.9975~m/s$ After the collision, the speed of the block and bullet is $0.9975~m/s$ We can find the time to fall 1.2 meters: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(1.2~m)}{9.80~m/s^2}}$ $t = 0.495~s$ We can find the horizontal distance the block moves in this time: $x = v_x~t = (0.9975~m/s)(0.495~s) = 0.494~m$ The block lands on the floor a distance of $0.494~m$ from the table.

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