Answer:
450 grams
Explanation:
Given:
mass of the bullet, m = 13.5 g = 0.0135 kg
velocity of the bullet, v = 253 m/s
spring constant of the spring, k = 205 N/m
Compression of the spring, x = 35.0 cm = 0.35 m
Now, the kinetic energy of the moving system (bullet + board) will tend to move the spring
thus,
let velocity of the system, V
now, applying the concept of conservation of momentum, we have
mv = (M + m)V
where,
M is the mass of the block
thus,
V = mv/(M + m)
now,
the kinetic energy of the system = (1/2)(M + m)V²
or
the kinetic energy of the system = (1/2)(M + m)(mv/(m + M))²
Energy gained by the spring = (1/2)kx²
now,
equating both the energies, we get
(1/2)(M + m)(mv/(m + M))² = (1/2)kx²
or
(mv)²/(m + M) = kx²
on substituting the values, we get
(0.0135 × 253)²/(0.0135 + M) = 205 × (0.35)²
or
11.66/(0.0135 + M) = 25.1125
or
M = 0.450 kg = 450 grams
