A 3-gram bullet is fired horizontally into a 10-kilogram block of wood suspended by a rope from the ceiling. The velocity of the bullet just

Question

A 3-gram bullet is fired horizontally into a 10-kilogram block of wood suspended by a rope from the ceiling. The velocity of the bullet just before the collision is 750 m/s. The block swings in an arc, rising a height h from its lowest position. What is h in millimeters? (Hint: You must apply conservation of momentum to the collision to find out how fast the block+bullet move together before you can use conservation of Energy to find h.)

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Calantha 1 week 2021-07-22T19:49:44+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-07-22T19:51:15+00:00

    Answer:

    2.58 mm

    Explanation:

    mass of bullet, m = 3 g = 0.003 kg

    mass of block M = 10 kg

    initial velocity of bullet, u = 750 m/s

    Let they move together after collision with velocity v.

    Use conservation of momentum

    m x u + 0 + ( m + M) x v

    0.003 x 750 = ( 10 + 0.003) x v

    v = 0.225 m/s

    Let the both reach at height h.

    use conservation of energy

    Kinetic energy at the bottom = Potential energy at height

    0.5 x ( M + m) x v² = ( M + m) x g x h

    0.5 x 0.225 x 0.225 = 9.8 x h

    h = 2.58 x 10^-3 m

    h = 2.58 mm

    0
    2021-07-22T19:51:17+00:00

    Answer:

    The height raised by the blcok is 2.5 mm.

    Explanation:

    Given that,

    Mass of the bullet, m_1=3\ g=0.003\ kg

    Initial speed of the bullet, u_1=750\ m/s

    Mass of the block of wood, m_2=10\ kg

    It was at rest, initial speed of the blockof wood, u_2=0

    The block swings in an arc, rising a height h from its lowest position. The momentum remains consered. Let V is the velocity of the block +bullet system together. So,

    m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{0.003\times 750+0}{(0.003+10)}\\\\V=0.224\ m/s

    Now using the conservation of energy to find the height h raised by the blck. So,

    \dfrac{1}{2}mV^2=mgh\\\\h=\dfrac{V^2}{2g}\\\\h=\dfrac{(0.224)^2}{2\times 10}\\\\h=0.0025\ m\\\\h=2.5\ mm

    So, the height raised by the blcok is 2.5 mm.

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