Center of MassHard
Question
A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for ᐃt = 0.01 sec., the impulse of the impact force is : (take g = 10 m/s2)
Options
A.1.3 N−s
B.1.06 N−s
C.1300 N−s
D.105 N−s
Solution
v1 = 
k2 = 
k1 ⇒ v22 = v12
∴ v2 =
= 5√2
|ᐃP| = |-mv2 - (mv1)| = m |-v2 - v1|
|ᐃP| = 50 × 10-3 ×
× 10 √2 = 
J = ᐃP = 1.05N-s.
k2 = k1 ⇒ v22 = v12 ∴ v2 =
= 5√2|ᐃP| = |-mv2 - (mv1)| = m |-v2 - v1|
|ᐃP| = 50 × 10-3 ×
× 10 √2 = J = ᐃP = 1.05N-s.
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