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Q.A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms⁻¹. If the mass of the ball is .15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball.)

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2019Subjective· 2mImportance★★★★★
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Impulse J=Δp=m(v−u)J = \Delta p = m(v-u); since the ball reverses direction at the same speed, ∣Δp∣=m×2u=0.15×24=3.6|\Delta p| = m \times 2u = 0.15 \times 24 = 3.6 N·s.

Take the direction from bowler to batsman as positive. Initial velocity of the ball, u=+12 m/su = +12\text{ m/s}. The batsman hits it straight back towards the bowler without changing its speed, so final velocity v=−12 m/sv = -12\text{ m/s}.

Mass of ball, m=0.15 kgm = 0.15\text{ kg}.

Impulse imparted equals the change in momentum:

J=Δp=m(v−u)=0.15×(−12−12)=0.15×(−24)=−3.6 kg⋅m/sJ = \Delta p = m(v - u) = 0.15 \times (-12 - 12) = 0.15 \times (-24) = -3.6\text{ kg·m/s}

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