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Vlad1618 [11]
4 years ago
9

The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhe

n it becomes embedded in the 10-kg block, determine thedistance the block will slide up along the plane before momentarily stopping
Physics
1 answer:
WARRIOR [948]4 years ago
4 0

Answer:

6.8 mm

Explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet,m_b=10 g=10\times 10^{-3} kg

1 kg=1000 g

Total mass of system,M=m+m_a=10+10\times 10^{-3}=10.01kg

Speed of bullet,u=300 m/s

\theta=30^{\circ}

By law of conservation of momentum

m_bucos\theta=Mv

v=\frac{m_bvcos\theta}{M}=\frac{0.01\times 300cos30^{\circ}}{10.01}=0.259m/s

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

\frac{1}{2}Mv^2-0=Mgh-0

\frac{1}{2}(10.01)(0.259)^2=10.01\times 9.8 h

Where g=9.8 m/s^2

h=\frac{(0.259)^2}{2\times 9.8}=0.0034m

1m=100 cm

h=0.0034\times 100=0.34 cm

Distance traveled by block=d=\frac{h}{sin\theta}=\frac{0.34}{sin30^{\circ}}=0.68 cm=6.8 mm

1cm=10 mm

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