a) A high speed bullet of mass 3.0 g and speed v strikes a block of mass 3.0 kg

Question

a)  A high speed bullet of mass 3.0 g and speed v strikes a block of mass 3.0 kg suspended from a light 1.0 m long string. The bullet becomes embedded in the block, which swings up so that the string makes an angle of 35 degrees with the vertical at its high point. How fast was the bullet going?

Answer: 1885 m/s

b)  A red ball of mass 2.0 kg has a velocity of 3i m/s. It elastically strikes a resting blue ball of mass 4.0 kg head on. The blue ball's final velocity is seen to be entirely in the i direction. What is the final velocity of the red ball? Hint: you need to use conservation of momentum and energy.

Answer:  -i  m/s

Explanation / Answer

m1 = 0.003 kg, m2 = 3 kg

height raised by pendulum

h = L(1-cos(35)) = 1*(1 - 0.819 ) = 0.18 m


let u is the speed os the bullet befor the collision.

and v is the speed of pendulum and bullet after the collision


0.5*(m1+m2)*v^2 = (m1+m2)*g*h


v = sqrt(2*g*h) =sqrt(2*9.8*0.18) = 1.8783 m/s

now, according to law of conservatio of momrntum

m1*u1 = (m1+m2)*v

u1 = (m1+m2)*v/m1 = (0.003+3)*1.8783/0.002 = 1880.2 m/s


b)
m1 = 2kg, u1 = 3m/s (red ball)
m2 = 4 kg, u2 = 0 (blue ball)

v1 = (m1-m2)*u1/(m1+m2) = (2-4)*3/(2+4) = -1 m/s

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