A bluet with mass of 200 gr hit a 4.80 kg block with speed of 400 m/s as it is s

Question

A bluet with mass of 200 gr hit a 4.80 kg block with speed of 400 m/s as it is shown in the figure below. The horizontal track is 28 m long and has a kinetics coefficient of friction mu_k=0.4. a) What would be the block's velocity at point C where the spring's length is equal to its natural length? b) How much does the block compress the spring (k=4.00 kN/m) before coming momentarily to rest? c) Where does the block ultimately come to rest? d) What would be the frequency of spring oscillation if we assume the block stick to the spring and oscillate on the frictionless surface?

Explanation / Answer

a) Momentum conservation for bullet and block B system

After hitting the block, the bullet will embed and both will move together

Mbullet = 0.2 kg

Ubullet = 400 m/s

MB = 4.8 kg

UB = 0

Msystem = Mbullet + MB = 0.2 + 4.8 = 5 kg

Mbullet*Ubullet + 0 = Msystem*Usystem

0.2*400 = 5*Usystem

Usystem = 16 m/s

System will move forward with deceleration de to friction

a = -ug = -0.4*9.8 = - 3.92 m/s^2

s = 28 m

Vsystem^2 = Usystem^2 + 2a*s

Vsystem^2 = 16^2 - 2*3.92*28 = 36.48

Vsystem = 6.03 m/s at point C

b) Now initial veolocity of the system = 6.03 m/s

force due to spring = -kx

x is the spring compression before system coming to rest

deceleration = -kx/Msystem = -800x

v^2 = u^2 + 2as

0 = 6.03^2 - 2*800x*x

x^2 =6.03^2 / 2*800 = 0.227

x = 0.15 m

c) Block comes to rest at a distance = 28 + 0.15 = 28.15 m from initial position

d) Frequency of oscillation f = sqrt (k/m) / (2 * pi) = sqrt(4000 / 5) / (2 * pi) = 4.5 Hz

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