Question

An 0.007 kg bullet is fired into a 0.23 kg block that is initially at rest at the edge of a table of height h = 1.05 meter. The bullet remains in the block, and after the impact the block lands 1.84 meters from the bottom of the table. What is the initial speed of the bullet?

Answer 1

Answer on Question #53378, Physics / Mechanics | Kinematics | Dynamics

An 0.007 kg bullet is fired into a 0.23 kg block that is initially at rest at the edge of a table of height h = 1.05 meter. The bullet remains in the block, and after the impact the block lands 1.84 meters from the bottom of the table. What is the initial speed of the bullet?

Solution:

The equation that denotes the conservation of momentum is:

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

where, $m_{1} =$ mass of body 1

$m_{2} =$ mass of object or body 2

$v_{1i} =$ initial velocity of body 1

$v_{2i} = 0$ initial velocity of body 2

$v_{f} =$ final velocity of both the objects

Thus,

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

Equations related to trajectory motion after the impact are given by

Horizontal distance, $d = v_{f}t$

Vertical distance, $y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2}$

At end of trajectory $y = 0$ .

Thus,

$$0 = h + 0 \cdot t - \frac {1}{2} g t ^ {2}$$ $$h = \frac {1}{2} g t ^ {2}$$ $$t = \sqrt {\frac {2 h}{g}}$$

So,

$$d = v _ {f} t = v _ {f} \sqrt {\frac {2 h}{g}}$$ $$v _ {f} = d \sqrt {\frac {g}{2 h}} = 1. 8 4 * \sqrt {\frac {9 . 8 1}{2 * 1 . 0 5}} = 3. 9 7 7 \mathrm {m / s}$$

The initial speed of bullet

$$v _ {1 i} = \frac {(m _ {1} + m _ {2}) v _ {f}}{m _ {1}} = \frac {(0 . 0 0 7 + 0 . 2 3) * 3 . 9 7 7}{0 . 0 0 7} \approx 1 3 5 \mathrm {m / s}$$

Answer: $135\mathrm{m / s}$

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