Respuesta :
Answer:
47.06 m/s
Explanation:
Complete statement of the question is
A bullet of mass 200 g travelling horizontally towards the east with speed 400 m/s, which strikes a block of mass 1.5 kg that is initially at rest on a frictionless table. After striking the block, the bullet is embedded in the block and the block and the bullet move together as one unit. What is the magnitude and direction of the velocity of the block/bullet combination immediately after the impact
[tex]m[/tex] = mass of the bullet = 200 g = 0.2 kg
[tex]M[/tex] = mass of the block = 1.5 kg
[tex]v[/tex] = speed of the bullet before collision = 400 m/s
[tex]V[/tex] = speed of the block before collision = 0 m/s
[tex]V'[/tex] = speed of the bullet/block combination after collision = ?
Using conservation of momentum
[tex]mv + MV = (m + M) V'\\(0.2) (400) + (1.5) (0) = (0.2 + 1.5) V'\\V' = 47.06[/tex]m/s
Answer:
[tex]v=\frac{mu}{M+m}[/tex]
Explanation:
Let the bullet is moving along the +X axis direction, as it strikes with the blcok which is at rest and embedded into it.
According to the conservation of momentum
m x u + M x 0 = (m + M) x v
where, m is the mass of bullet, M is the mass of block, u be the initial velocity of the bullet and v be the velocity of combined mass,
Here we observe that the direction of v is same as the direction of u.
[tex]v=\frac{mu}{M+m}[/tex]
