Question

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/swhen it becomes embedded in the 10-kg block, determine thedistance the block will slide up along the plane before momentarily stopping

Answer 1

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