Question

A 5.00-g bullet moving with an initial speed of 450 m/s passes through a 1.00-kg block, as shown in the figure. The block is initially at rest on a frictionless, horizontal surface and is connected to a spring that has a spring-force constant of 900 N/m. The block moves a distance d = 3.34 cm to the right after the collision. Note that the block only starts moving after the bullet has completely passed through the block. IWW d Round your answers to 3 significant figures. • What is the velocity, in m/s, of the block when it starts moving after being impacted by the bullet? Choose... • What is the speed of the bullet, in m/s, after it passes through the block? Choose... • What is the magnitude of the change in the total kinetic energy, in l, after the collision? Choose... " Choose... 1.50 2.00 300 350 250

Answer 1

here,

the mass of bullet , m1 = 5 g = 0.005 kg

the mass of block , m2 = 1 kg

the initial velocity of bullet , u1 = 450 m/s

the force constant , K = 900 N/m

d = 3.34 cm = 0.0334 m

let the velocity of the block be v

using conservation of energy

0.5 * K * d^2 = 0.5 * (m2) * v^2

0.5 * 900 * 0.0334^2 = 0.5 * ( 1) * v^2

solving for v

v = 1 m/s

the speed of block after the collison is 1 m/s

let the speed of bullet be v1

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

0.005 * 450 = 0.005 * v1 + 1 * 1

v1 = 250 m/s

the speed of bullet is 250 m/s

the magnitude of change in energy , U = 0.5 * (m2 * v2^2 + m1 * v1^2 - m1 * u1^2 )

U = 0.5 * ( 1 * 1^2 - 0.005 * 250^2 - 0.005 * 450^2) J

U = - 349.5 J

the magnitude of change in total energy is 349.5 J