Respuesta :
Answer:
Combined velocity is 402.35 m/s.
Explanation:
Given that,
Mass of the bullet, m = 0.045 kg
Initial speed of the bullet, v = 456 m/s
Mass of the sponge, m' = 0.006 kg
Initial speed of the sponge is 0 as it was at rest. The momentum remains conserved in this case. The bullet is embedded in the sponge, it is a case of inelastic collision. So,
[tex]mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.045\times 456+0}{(0.045+0.006)}\\\\V=402.35\ m/s[/tex]
So, the combined bullet and sponge velocity after the perfectly inelastic collision is 402.35 m/s.
Answer:
Explanation:
To find the velocity of combined body we shall apply conservation of momentum
initial momentum = .045 x 456 = 20.52 kg m /s
mass of combined body = .045 +.006
= .051 kg
If v be the velocity of combined mass
.051 v = 20.52
v = 20.52 / .051
= 402.35 m/s
If h be the heights up to which combined mass go with deceleration g
v² = 2gh
h = v² / 2g
= 402.35² / 2 x 9.8
= 8259.46 m
