M2 Jan 03 - Q7 (Vectors)

A ball B of mass 0.4 kg is struck by a bat at a point O which is 1.2 m above horizontal ground. The unit vectors i and j are respectively horizontal and vertical. Immediately before being struck, B has velocity (-20i + 4j) ms-1. Immediately after being struck it has velocity (15i + 16j) ms-1.

After B has been struck, it moves freely under gravity and strikes the ground at the point A, as shown in Fig. 3. The ball is modelled as a particle.

(a) Calculate the magnitude of the impulse exerted by the bat on B.

(b) By using the principle of conservation of energy, or otherwise, find the speed of B when it reaches A.

I am stuck on B, i have the mark scheme but i have no idea where the answer is comming from.

Answer to A is needed is 14.8Ns

Thanks

Reply 1

meef cheese

Total energy at before = total energy at point A

Before: PE = mgh = (0.4)(9.8)(1.2) = 4.704 J

KE = 0.5mv² = 0.5(0.4)(15² + 16²) = 96.2 J

Total Energy = 100.904 J

At B: PE = 0

KE = 100.904
=> 0.5(0.4)v² = 100.904
=>0.2 v² = 1009.4
=> v² = 504.52
=> v = 22.46 m/s

Reply 2

sshum

Thank you so much,
I understand it now
+rep

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