1. A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest.Calculate velocity of ball of mass 6kg after collision. 2. A 10.0g bullet is fired into a stationary block of wood (m=5.00kg) . The bullet sticks into the block, and the speed of the bullet-phis combination immedately after collision is 0.600m/s What was the original speed of the bullet? 3. A block of mass m_(1)=1.6kg initially moving to the right with a speed of 4m/s on a horizontal frictionless track collides with a block of mass m_(2)=2.1kg initially moving to the left with speed of 2.5m/s If the collision is elastic, find the velocities of the two block after collision? 4. A partcle of mass 4.Okg initially moving with velocity of 2.0m/s collides with a partcle of mass 6.Okg, initially moving velocity of -4m/s What are the velocity of the two particle after collision? 5. A 4kg block moving right at 6m/s collides elastically with a 2kg moving at 3m/s left, find final velocities the blocks.

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1. A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest.Calculate velocity of ball of mass 6kg after collision. 2. A 10.0g bullet is fired into a stationary block of wood (m=5.00kg) . The bullet sticks into the block, and the speed of the bullet-phis combination immedately after collision is 0.600m/s What was the original speed of the bullet? 3. A block of mass m_(1)=1.6kg initially moving to the right with a speed of 4m/s on a horizontal frictionless track collides with a block of mass m_(2)=2.1kg initially moving to the left with speed of 2.5m/s If the collision is elastic, find the velocities of the two block after collision? 4. A partcle of mass 4.Okg initially moving with velocity of 2.0m/s collides with a partcle of mass 6.Okg, initially moving velocity of -4m/s What are the velocity of the two particle after collision? 5. A 4kg block moving right at 6m/s collides elastically with a 2kg moving at 3m/s left, find final velocities the blocks.

Answer 1

### 1. $4 \text{ m/s}$ ### 2. $600 \text{ m/s}$ ### 3. $v_{1f} = 1.25 \text{ m/s}, v_{2f} = -0.5 \text{ m/s}$ ### 4. $v_f = -1.6 \text{ m/s}$ ### 5. $v_{1f} = 1 \text{ m/s}, v_{2f} = 8 \text{ m/s}$

Answer 2

## Step1: Identify Conservation Laws ### For all the problems, apply conservation of momentum and energy (where applicable). ## Step2: Use Momentum Conservation Formula ### Apply $\sum m_1 v_{1i} + \sum m_2 v_{2i} = \sum m_1 v_{1f} + \sum m_2 v_{2f}$ before and after the collision. # Problem 1: ## Step3: Initial and Final Parameters ### Mass of moving ball $m_1 = 2 \text{ kg}$, $v_{1i} = 12 \text{ m/s}$, $v_{1f} = 0 \text{ m/s}$ and stationary ball $m_2 = 6 \text{ kg}$, $v_2 = 0$. Use conservation of momentum. $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$ $$2 \cdot 12 + 6 \cdot 0 = 2 \cdot 0 + 6 \cdot v_{2f} $$ $$24 = 6v_{2f} $$ ## Step4: Solve for $v_{2f}$ ### $$v_{2f} = 4 \text{ m/s}$$ # Problem 2: ## Step3: Initial and Final Parameters ### Bullet mass $m_1 = 0.01 \text{ kg}$, and block mass $m_2 = 5 \text{ kg}$, combined velocity $v_f = 0.6 \text{ m/s}$. Use conservation of momentum: $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f$$ $$0.01 v_{1i} + 5 \cdot 0 = (0.01 + 5) \cdot 0.6$$ ## Step4: Solve for $v_{1i}$ ### $$0.01 v_{1i} = 3.006$$ $$v_{1i} = 600 \text{ m/s}$$ # Problem 3: ## Step3: Initial and Final Parameters ### Using elastic collision for masses $m_1 = 1.6 \text{ kg}$, $v_{1i} = 4 \text{ m/s}$, $v_{1f}$ unknown; $m_2 = 2.1 \text{ kg}$, $v_{2i} = -2.5 \text{ m/s}$, $v_{2f}$ unknown. Apply conservation of momentum: $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$ $$1.6 \cdot 4 + 2.1 \cdot -2.5 = 1.6 v_{1f} + 2.1 v_{2f}$$ $$6.4 - 5.25 = 1.6 v_{1f} + 2.1 v_{2f}$$ ## Step4: Solve Simultaneous Equations ### Apply: $$v_{1i} - v_{2i} = v_{2f} - v_{1f}$$ $$4 - (-2.5) = v_{2f} - v_{1f}$$ ## Final Velocities: ### $(v_{1f}, v_{2f}) = (1.25, -0.5)$ (Further solved using substitution or systems of equations method) # Problem 4: ## Step3: Initial and Final Parameters ### Completely inelastic collision: masses $m_1 = 4 \text{ kg}$, $v_{1i} = 2 \text{ m/s}$, mass $m_2 = 6 \text{ kg}$, $v_2 = -4 \text{ m/s}$. Use conservation of momentum: $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f$$ ## Step4: Solve for $v_f$ ### $$4 \cdot 2 + 6 \cdot (-4) = (4+6)v_f$$ $$8 - 24 = 10v_f$$ $$v_f = -1.6 \text{ m/s}$$ # Problem 5: ## Step3: Initial and Final Parameters ### Using elastic collision, $m_1 = 4 \text{ kg}$, $v_{1i} = 6 \text{ m/s}$, $v_{1f}$ unknown; $m_2 = 2 \text{ kg}$, $v_{2i} = -3 \text{ m/s}$, $v_{2f}$ unknown. Apply conservation of momentum: $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$ $$4 \cdot 6 + 2 \cdot -3 = 4 v_{1f} + 2 v_{2f}$$ $$24 - 6 = 4 v_{1f} + 2 v_{2f}$$ ## Step4: Solve Simultaneous Equations ### Further applying relative velocity: $$v_{1i} - v_{2i} = v_{2f} - v_{1f}$$ $$6 - (-3) = v_{2f} - v_{1f}$$ ## Final Velocities: ### $(v_{1f}, v_{2f}) = 1 \text{ m/s}, 8 \text{ m/s}$ (Further solved using substitution or systems of equations method)

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