An 8.0-g bullet is suddenly shot into a 4.0-kg block, at rest on a frictionless

Question

An 8.0-g bullet is suddenly shot into a 4.0-kg block, at rest on a frictionless horizontal surface, as shown in the figure. The bullet remains lodged in the block. The block then moves against a spring and compresses it by 8.9 cm. The force constant (spring constant) of the spring is 1400 N/m. What is the magnitude of the impulse on the block (including the bullet inside) due to the spring during the entire time interval during which the block is in contact with the spring? Please show your own work.

answers are :

6.7 N • s   

8.3 N • s

10 N • s

12 N • s

13 N • s

Explanation / Answer

When bullet shot into the block then its kinetic energy will restore in the form of potential energy of the spring.
Kinetic energy of the block and bullet = (1/2)(M+m)V2
Where M is the mass of block and m is the mass of bullet.
V is the velocity of the block and bullet.
This kinetic energy will restore into the potential energy of the spring.
Potential energy of the spring = (1/2)kx2
where k is spring constant and x is the compression of spring
Potential energy of the spring = (1/2)*(1400)*(0.089)2
Hence by the energy conservation
(1/2)(M+m)V2 = (1/2)*(1400)*(0.089)2
V = 1.663 m/s
Hence the initial momentum of the block and bullet = (M+m)V = 6.67 kgm/s
Final momentum of the block will be zero because it will come to rest by giving its all energy to the spring.
We know that the
Impulse = Change in momentum
Impule = Initial momentum - Final momentum
   = 6.67 - 0
Impulse = 6.67 kgm/s or Ns

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