A 700-g block of ice is initially at rest on a frictionless surface. A 70-g steel ball traveling at 4 m/sec strikes the block of ice and bounces backward at 1.5 m/sec. How fast will the block of ice be moving after the collision?

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0.35 m/sec. Answer: 0.35 m/sec. Use the conservation of momentum equation to solve for the speed of the block of ice. $$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$ Where: $$m_1$$ is the mass of the steel ball, $$v_1$$ is the initial velocity of the steel ball, $$v_2$$ is the initial velocity of the block of ice (which is 0), $$v_1'$$ is the final velocity of the steel ball, and $$v_2'$$ is the final velocity of the block of ice. $$m_1 = 70 	ext{ g}, \ v_1 = 4 	ext{ m/sec}, \ v_2 = 0 	ext{ m/sec}, \ v_1' = -1.5 	ext{ m/sec}$$ Solving for $$v_2'$$, we get: $$v_2' = \frac{(70 	ext{ g} 	imes 4 	ext{ m/sec}) + (700 	ext{ g} 	imes 0 	ext{ m/sec}) - (70 	ext{ g} 	imes -1.5 	ext{ m/sec})}{700 	ext{ g}} = 0.35 	ext{ m/sec}$$

A 700-g block of ice is initially at rest on a frictionless surface. A 70-g steel ball traveling at 4 m/sec strikes the block of ice and bounces backward at 1.5 m/sec. How fast will the block of ice be moving after the collision?

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