a bullet of mass 20g moving with velocity of 300 m/s gets embedded in a truly suspended wooden block of mass 800g. calculate the velocity aquired by the block?
Answers
Answer:
This is completely inelastic collision. The linear momentum is conserved.
If a body of mass, m collides with a body of mass M and gets embedded in it, according to law of conservation of momentum,
mv =( m+M)V.
Here,
m is mass of colliding body. m= 20 g.
M is mass of target body. M=880 g.
v=300m/s is velocity of colliding body.
V= ? V is velocity of combined system of m and M.
Therefore 20(g).300(m/s)=(880+20)( g)V(m/s.)OR
V=6000/900=6.67 m/s.
Explanation:
hope it's helpful for you
Answer: This is completely inelastic collision. The linear momentum is conserved.
If a body of mass, m collides with a body of mass M and gets embedded in it, according to law of conservation of momentum,
mv =( m+M)V.
Here,
m is mass of colliding body. m= 20 g.
M is mass of target body. M=880 g.
v=300m/s is velocity of colliding body.
V= ? V is velocity of combined system of m and M.
Therefore 20(g).300(m/s)=(880+20)( g)V(m/s.)OR
V=6000/900=6.67 m/s.