A bullet of mass 24g travelling in a horizontal path with a velocity of 450ms-1 strikes a wooden block of wood of mass 976g resting...

      

A bullet of mass 24g travelling in a horizontal path with a velocity of 450ms-1
strikes a wooden block of wood of mass 976g
resting on a rough horizontal surface. After impact, the bullet and the block move together for a distance of 7.5m before
coming rest.
(a) Name the type of collision which takes place above (1 mark)
(b) What‘s the velocity of the two bodies when they start sliding (2 marks)
(c) Calculate the force which brings the two bodies to rest (3 marks)
(d) Determine the coefficient of friction between the block and the surface during this motion

  

Answers


john
(a) Inelastic collision
(b) m1u1 + m2u2 = (m1 +m2)v
22/1000X450 + 976/1000X 0= (976/1000 +
26/1000)v
v =10.8ms-1
(c) V
2
= u
2
+ 2 a s
0= 10.82
+ 2 X a X 7.5
a = - 10.82
/15
a =- 7.7776 ms-1
F = ma = 1 X -7.7776 ms-1
F = -7.7776 N
john3 answered the question on November 2, 2017 at 21:04


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