What is the final velocity of the lighter object? (See below)

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What is the final velocity of the lighter object? (See below)

Two objects whose masses are $1 k g$ $1 kg$ and $4 k g$ $4 kg$ collide with equal and opposite velocities of magnitude $10 m^{- 1}$ $10 m^-1$ . They bounce off one another, and the heavier object has a final velocity of $5 m s^{- 1}$ $5ms^-1$ in its original direction. What is the final velocity of the lighter object?

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Answer 1 · 2018-05-14T18:01:56

The problem is incorrectly specified. The heavier object continues on at $- 2 m/s$ $-2" m/s"$ and the lighter object rebounds at $- 22 m/s$ $-22" m/s"$

Explanation:

Given: $m_{1} = 1 kg, m_{2} = 4 kg, \to u_{1} = 10 m/s, \to u_{2} = - 10 m/s, \to v_{2} = - 5 m/s$ $m_1 = 1" kg", m_2 = 4" kg", vec(u_1)= 10" m/s", vec(u_2) = -10" m/s", vec(v_2) = -5" m/s"$

NOTE: I have chosen the direction of the velocity of the lighter object to be positive and the direction of the velocity of the heavier object to be negative.

Take the following equation from this Momentum reference to check whether the initial information is consistent:

$\to v_{2} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} \to u_{2} + \frac{2 m_{1}}{m_{1} + m_{2}} \to u_{1}$ $vec(v_2) = (m_2-m_1)/(m_1+m_2)vec(u_2)+ (2m_1)/(m_1+m_2)vec(u_1)$

We know all of these values:

$\to v_{2} = \frac{4 kg - 1 kg}{1 kg + 4 kg} (- 10 m/s) + \frac{2 (1 kg)}{1 kg + 4 kg} (10 m/s)$ $vec(v_2) = (4" kg"-1" kg")/(1" kg"+4" kg")(-10" m/s")+ (2(1" kg"))/(1" kg"+4" kg")(10" m/s")$

$\to v_{2} = - 2 m/s \leftarrow This is inconsistent$ $vec(v_2) = -2" m/s" color(red)(larr "This is inconsistent"$

Let's find the velocity of $m_{1}$ $m_1$ :

$\to v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \to u_{1} + \frac{2 m_{2}}{m_{1} + m_{2}} \to u_{2}$ $vec(v_1) = (m_1-m_2)/(m_1+m_2)vec(u_1)+ (2m_2)/(m_1+m_2)vec(u_2)$

$\to v_{1} = \frac{1 kg - 4 kg}{1 kg + 4 kg} (10 m/s) + \frac{2 (4 kg)}{1 kg + 4 kg} (- 10 m/s)$ $vec(v_1) = (1" kg"-4" kg")/(1" kg"+4" kg")(10" m/s")+ (2(4" kg"))/(1" kg"+4" kg")(-10" m/s")$

$\to v_{1} = - 22 m/s$ $vec(v_1) = -22" m/s"$

Answer 2 · 2018-05-14T18:02:32

10 m/s in the direction opposite its original motion.

Explanation:

$Momentum before = momentum after$ $"Momentum before = momentum after"$

$m_{1} \cdot u_{1} + m_{2} \cdot u_{2} = m_{1} \cdot v_{1} + m_{2} \cdot v_{2}$ $m_1*u_1 + m_2*u_2 = m_1*v_1 + m_2*v_2$

$1 kg \cdot (- 10 \frac{m}{s}) + 4 kg \cdot 10 \frac{m}{s} = 1 kg \cdot v_{1} + 4 kg \cdot 5 \frac{m}{s}$ $1 "kg"*(-10 m/s) + 4 "kg"*10 m/s = 1 "kg"*v_1 + 4 "kg"*5 m/s$

Solving for $v_{1}$ $v_1$

$v_{1} = \frac{1 kg \cdot (- 10 \frac{m}{s}) + 4 kg \cdot 10 \frac{m}{s} - 4 kg \cdot 5 \frac{m}{s}}{1 k g}$ $v_1 = (1 "kg"*(-10 m/s) + 4 "kg"*10 m/s - 4 "kg"*5 m/s)/(1 kg)$

$v_{1} = \frac{- 10 kg \cdot \frac{m}{s} + 40 kg \cdot \frac{m}{s} - 20 kg \frac{m}{s}}{1 k g}$ $v_1 = (-10 "kg"*m/s + 40 "kg"*m/s - 20 "kg"m/s)/(1 kg)$

$v_1 = (10 cancel(kg)*m/s)/(1 cancel(kg)) = 10 m/s$

Since $v_1$ is positive but $u_1$ was negative, this is the opposite direction of its original motion.

I hope this helps,
Steve

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What is the final velocity of the lighter object? (See below)

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