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Extra resources for Applied Mathematics by Example: Exercises

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5 metres away, returning along the same path. 5. How far will A have travelled before B hits it again? 18. 1 m/s, strikes the cushion at an angle of 30◦ . 8. Calculate (a) the angle at which it rebounds, (b) its speed after the impact and (c) the impulse it exerts on the cushion. 19. A cricket ball travelling at 25 m/s hits the pitch at an angle of 17◦ to the horizontal. 7, at what speed and in what direction is it travelling immediately after it has bounced? 6 m high and a horizontal distance of 3 m from the point where the ball bounces?

Distance covered = 3 km = area = 1/2 × base × height = 1/2 × (5/60) × Vmax , Vmax = 72 km/hr. We do not know where the apex of the triangle is, but we can still use the area formula 1/2 × base × height. velocity (km/hr) Vmax 3 km 5 t (mins) 12. Distance A covers in first 2 seconds = 1/2 × 6 × 22 = 12 m. 5 × 12 = 30 m. 295 s to cover 100 m. 5 m. e. 5 seconds when speeds of both are 11 m/s. 5 m behind. 301 s. 301 s, so A wins. 006 s × 10 m/s = 6 cm in distance. 2 Projectiles 1. 476 s. 11 m. 14 m.

What is his velocity vector? What are the position vectors of P, Q and S, 5 seconds after P and Q have started running? Will either Q or S succeed in intercepting P? 23. A particle P has velocity 3i + 4j m/s, where i and j are unit vectors along the xand y-axes respectively. What is the speed of P? What is the angle between its direction of motion and (a) the x-axis and (b) the y-axis? P receives an impulse in a collision after which its velocity becomes 5i + 4j m/s. What is (i) the change in velocity of P, (ii) the increase in speed and (iii) the angle through which its direction of motion has been deflected?

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