Please, if someone can help me with finding these answers. Thank you. 2) A 3.20
ID: 1465549 • Letter: P
Question
Please, if someone can help me with finding these answers. Thank you.
2) A 3.20 kg object initially moving in the positive x-direction with a velocity of +5.11 m/s collides with and sticks to a 1.84 kg object initially moving in the negative y-direction with a velocity of -2.56 m/s. Find the final components of velocity of the composite object. (Indicate the direction with the sign of your answer.)
vfx= _____m/s
vfy=______ m/s
(a) Find the initial speed of the bullet–block system.
=______ m/s
(b) Find the speed of the bullet.
=______ m/s
vfx= _____m/s
vfy=______ m/s
Explanation / Answer
2) m1 =3.2 kg , u1x =5.11 m/s, m2 =1.84 kg, vy =-12.56 m/s
From conservation of momentum along x-direction
m1u1x +0 =(m1+m2)vfx
3.2*5.11 = (3.2+1.84)vfx
vfx = 3.24 m/s
From conservation of momentum along x-direction
0 +m2u2y =(m1+m2)vfy
-1.84*2.56 = (3.2+1.84)vfy
vfy = -0.935 m/s
3) m =4.5 g , M =2.098 kg , k =5.75xx10^2 N/m , x =6.44 cm
M+m = 0.0045+2.098= 2.1025 kg
(a) From conservation of energy
(1/2)(m+M)v^2 =(1/2)kx^2
v = [k/(M+m)]^1/2 (x) =(0.0644)[575/2.1025]^1/2
speed of bullet -block system v = 1.065 m/s
(b) From conservation of momentum
mu +0 = (m+M)v
u = (2.1025*1.065)/(0.0045)
Speed of bullet u = 497.6 m/s