An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 7.9 cm. The force constant of the spring is 1400 N/m. What is the impulse of the block (including the bullet), due to the spring, during the entire time interval in which block and spring?
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6.65Ns
To find the impulse of the block (including the bullet) due to the spring, we need to calculate the change in momentum. The impulse-momentum theorem states that the impulse acting on an object is equal to the change in momentum of that object.
First, let's calculate the initial momentum of the block and bullet system. Momentum (p) is given by the product of mass (m) and velocity (v):
Initial momentum = m * v
The bullet has a mass of 8 g, which is 0.008 kg. The block has a mass of 4.0 kg. Before the collision, the bullet and block are at rest, so the initial velocity is 0 m/s.
Therefore, the initial momentum is:
Initial momentum = (0.008 kg + 4.0 kg) * 0 m/s = 0 kg·m/s
Next, let's calculate the final momentum of the block and bullet system after the block compresses the spring. The bullet remains lodged in the block, so their combined mass is now 0.008 kg + 4.0 kg = 4.008 kg.
To calculate the final velocity (vf), we can use the conservation of momentum. Since there is no external force acting horizontally on the system, the momentum before the collision equals the momentum after the collision:
Initial momentum = Final momentum
0 kg·m/s = (4.008 kg) * vf
Solving for vf:
vf = 0 m/s
Since the final velocity is 0 m/s, there is no change in momentum. Therefore, the change in momentum is:
Change in momentum = Final momentum - Initial momentum = 0 kg·m/s - 0 kg·m/s = 0 kg·m/s
Finally, we can calculate the impulse using the formula:
Impulse = Change in momentum
Impulse = 0 kg·m/s
Therefore, the impulse of the block (including the bullet) due to the spring during the entire time interval is 0 kg·m/s.