A block of mass 10 text{ g} is suspended from | vedclass

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(D) Step $1$: Apply the law of conservation of linear momentum during the collision.Let $m_1 = 10 	ext{ g} = 0.01 	ext{ kg}$ be the mass of the bull

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(D) Step $1$: Apply the law of conservation of linear momentum during the collision.Let $m_1 = 10 	ext{ g} = 0.01 	ext{ kg}$ be the mass of the bullet and $m_2 = 10 	ext{ g} = 0.01 	ext{ kg}$ be the mass of the block.Initial velocity of the bullet $u_1 = 1 	ext{ m/s}$ and initial velocity of the block $u_2 = 0 	ext{ m/s}$.Let $v$ be the common velocity of the system (bullet + block) immediately after the collision.$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$$0.01 	imes 1 + 0.01 	imes 0 = (0.01 + 0.01) v$$0.01 = 0.02 v$$v = 0.5 	ext{ m/s}$.Step $2$: Apply the law of conservation of mechanical energy to find the height $h$.The kinetic energy of the combined system is converted into potential energy at the maximum height $h$.$\frac{1}{2} (m_1 + m_2) v^2 = (m_1 + m_2) g h$$h = \frac{v^2}{2g}$$h = \frac{(0.5)^2}{2 	imes 10} = \frac{0.25}{20} = 0.0125 	ext{ m}$.Step $3$: Convert the height into centimeters.$h = 0.0125 	ext{ m} = 0.0125 	imes 100 	ext{ cm} = 1.25 	ext{ cm}$.

$A$ block of mass $10 \text{ g}$ is suspended from the ceiling by a string. $A$ bullet of mass $10 \text{ g}$ moving with a velocity of $1 \text{ m/s}$ strikes the block and gets embedded in it. To what height will the block rise (in $\text{ cm}$)? (Take $g = 10 \text{ m/s}^2$)

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