May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
(If the velocities are not along the same direction, (As the ball returns to its initial position, the change→ → →→| s1+s2| | snet| (a) velocity of the particle after 2 sec.average velocity = VaV= =s1s2t(b) angle between initial velocity and the velocity after 2 sec.+netV1V(c) the maximum height reached by the projectile2(d) horizontal range of the projectile→ →Since s 1 = s 2 = d and s net = s + s | = 0you'll need to use the vector from of this equations,derived later in this section.) It's important to note thein position, the change in position vector of the ball,that is the net displacement will be zero).order of the double subscripts in equation (1) v A/B→always means "velocity of A relative to B." These ∴ | VaV| = 0.subscripts obey an interesting kind of algebra, asequation (1) shown. If regard each one as a fraction,2. A long belt is moving horizontally with a speed of 4then the fraction on the left side is the product of theKm/hour. A child runs on this belt to and fro with afractions on the right sides : P/A = (P/B) (B/A). Thisspeed of 9 Km/hour (with respect to the belt) betweenis a handy rule you can use when applying Equationhis father and mother located 50 m apart on the(1) to any number of frames of reference. Formoving belt. For an observer on a stationary platformexample, if there are three different frames ofoutside, what is thereference A, B, and C, we can write immediately. (a) speed of the child running in the direction of motionv P/A = v P/C + v C/B + V B/Aof the belt,Step 4 : Evaluate your answer : Be on the lookout forstray minus signs in your answer. If the target(b) speed of the child running opposite to the direction ofmotion of the belt andvariable is the velocity of a car relative to a bus(v V/B ), make sure that you haven't accidentallycalculated the velocity of the bus relative of the car(c) time taken by the child in case (a) and (b) ?Which of the answers change, if motion is viewed byone of the parents ?(v B/C ). If you have made this mistake, you canrecover using equation.v A/B = – v B/ASol. Let us consider positive direction of x-axis from leftto right(a) Here, v B = + 4 Km/hourSpeed of child w.r.t. belt, v C = = 9 Km/hourSolved Examples∴ Speed of child w.r.t. stationary observer,v C ′ = v C + v B or v C ′ = 9 + 4 = 13 Km/hour1. A small glass ball is pushed with a speed V from A.It moves on a smooth surface and collides with thewall at B. If it loses half of its speed during thecollision, find the distance, average speed andvelocity of the ball till it reaches at its initial position.(b) Here, v B = + 4 Km/hour, v C = – 9 Km/hour∴ Speed of child w.r.t. stationary observer,v C ′ = v C + v B or v C ′ = – 9 + 4 = –5 Km/hourThe negative sign shows that the child appears to runin a direction opposite to the direction of motion ofthe belt.A V 0.5V B(c) Distance between the parents, s = 50 m = 0.05 KmSince parents and child are located on the same belt,the speed of the child as observe by stationarydSol. The ball moves from A to B with a constant speed V.Since it loses half of its speed on collision, it returnsobserver in either direction (either father to mother orfrom mother to father) will be 9 Km/hour.Time taken by the child in case (a) and (b),from B to A with a constant speed V/2.0.50 km∴ V 1 = V and V 2 = V/2t = = 20 sec.9 km / hourd1+ d2Using the formula, V aV =If the motion is observed by one of parents, answer to(d1/ V 1)+ (d2/ V2)case (a) case (b) gets altred. It is because the speedPutting d 1 = d 2 = d; V 1 = V and V 2 = V/2of the child w.r.t. either of mother or father isd1+ d22V9 Km/hour.We obtain, V aV ==(dV) + (d / 0.5V) 33. A particle is projected with velocity v 0 = 100 m/s atFrom the formula,an angle θ = 30º with the horizontal. Find :| 1 2XtraEdge for IIT-JEE 28 MAY 2011
Sol. (a)(b)(c)→vt→= vxt→î + vytĵwhere î and ĵ are the unit vectors along +ve x and+ve y-axis respectively→tv =(u x + a x t) î + (u y + a y t) ĵHere, u x = v 0 cos θ = 50 3 m/s, a x = 0u y = v 0 sin θ = 50 m/s, a y = – g(Q g acts downwards)→tv = 503 î + (50 – 10 × 2) ĵ=[50 3 î + 30 ĵ ] m/s∴ | → 2 2v 2 | = (vx + v y ) = 2 2( 50 3) + (30)∴→0→2→0v = 50v = 503 î + 50 ĵ3 î + 30 ĵv . v → 2 = 7500 + 1500 = 9000If α is the angle between v → 0 and v→ 2→v0.v29000Then, cos α = =→ →100×91.65| v0| × | v2|α = cos –1 (0.98) = 10.8ºv 2 y – u 2 y = 2a y yAt y = y max , v y = 0∴ 0 – v 2 0 sin 2 θ = 2 (–g)y max2 2v0 sin θ∴ y max = = 125 m2g→u 2 sin 2θ (d) R = = 1732 mg4. A ball starts falling with zero initial velocity on asmooth inclined plane forming an angle α with thehorizontal. Having fallen the distance 'h', the ballrebounds elastically off the inclined plane. At whatdistance from the impact point will the ball reboundfor the second time ?αSol. Just before impact magnitude of velocity of the ball,v = ( 2gh)ααAs the ball collides elastically and the inclined planeis fixed, the ball follows the law of reflection.Now along the incline, velocity component afterimpact is v sin α and acceleration is g sin α.Perpendicular to the incline, velocity component isvcos α and acceleration (– g cos α). Hence, if wemeasure x and y-coordinate along the incline andperpendicular to the incline, thenx = (v sin α) t + ½ (g sin α)t 2and y = (v cos α) t – ½ (g cos α)t 2When the ball hits the plane for a second time,y = 0, (v cos α)t – ½(g cos α)t 2 or t = (2v/g)Putting this value of t in x,4v2 sin αx = = 8h sin αg5. A batsman hits a ball at a height of 1.22m above theground so that ball leaves the bat at an angle 45º withthe horizontal. A 7.31 m high wall is situated at adistance of 97.53 m from the position of the batsman.Will the ball clear the wall if its range is 106.68 m.Take g = 10 m/s 2v02 sin 2θSol. R(range) =gor,1.22m2 Rgv 0 = = Rg as θ = 45ºsin 2θA45ºv 0106.68mor, v 0 = ( Rg)…(1)Equation of trajectoryor, y = x –y = x tan 45º –2vgx 22Rg.½20gx2cos2gx 2= x – RgB45ºPutting x = 97.53, we get210×(97.53)y = 97.53 –= 8.35 cm106.68×10Hence, height of the ball from the ground level ish = 8.35 + 1.22 = 9.577 mAs height of the wall is 7.31 m so the ball will clearthe wall.XtraEdge for IIT-JEE 29 MAY 2011
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Sol. (a)(b)(c)→vt→= vxt→î + vytĵwhere î and ĵ are the unit vectors along +ve x and+ve y-axis respectively→tv =(u x + a x t) î + (u y + a y t) ĵHere, u x = v 0 cos θ = 50 3 m/s, a x = 0u y = v 0 sin θ = 50 m/s, a y = – g(Q g acts downwards)→tv = 503 î + (50 – 10 × 2) ĵ=[50 3 î + 30 ĵ ] m/s∴ | → 2 2v 2 | = (vx + v y ) = 2 2( 50 3) + (30)∴→0→2→0v = 50v = 503 î + 50 ĵ3 î + 30 ĵv . v → 2 = 7500 + 1500 = 9000If α is the angle between v → 0 and v→ 2→v0.v29000Then, cos α = =→ →100×91.65| v0| × | v2|α = cos –1 (0.98) = 10.8ºv 2 y – u 2 y = 2a y yAt y = y max , v y = 0∴ 0 – v 2 0 sin 2 θ = 2 (–g)y max2 2v0 sin θ∴ y max = = 125 m2g→u 2 sin 2θ (d) R = = 1732 mg4. A ball starts falling with zero initial velocity on asmooth inclined plane forming an angle α with thehorizontal. Having fallen the distance 'h', the ballrebounds elastically off the inclined plane. At whatdistance from the impact point will the ball reboundfor the second time ?αSol. Just before impact magnitude of velocity of the ball,v = ( 2gh)ααAs the ball collides elastically and the inclined planeis fixed, the ball follows the law of reflection.Now along the incline, velocity component afterimpact is v sin α and acceleration is g sin α.Perpendicular to the incline, velocity component isvcos α and acceleration (– g cos α). Hence, if wemeasure x and y-coordinate along the incline andperpendicular to the incline, thenx = (v sin α) t + ½ (g sin α)t 2and y = (v cos α) t – ½ (g cos α)t 2When the ball hits the plane for a second time,y = 0, (v cos α)t – ½(g cos α)t 2 or t = (2v/g)Putting this value of t in x,4v2 sin αx = = 8h sin αg5. A batsman hits a ball at a height of 1.22m above theground so that ball leaves the bat at an angle 45º withthe horizontal. A 7.31 m high wall is situated at adistance of 97.53 m from the position of the batsman.Will the ball clear the wall if its range is 106.68 m.Take g = 10 m/s 2v02 sin 2θSol. R(range) =gor,1.22m2 Rgv 0 = = Rg as θ = 45ºsin 2θA45ºv 0106.68mor, v 0 = ( Rg)…(1)Equation of trajectoryor, y = x –y = x tan 45º –2vgx 22Rg.½20gx2cos2gx 2= x – RgB45ºPutting x = 97.53, we get210×(97.53)y = 97.53 –= 8.35 cm106.68×10Hence, height of the ball from the ground level ish = 8.35 + 1.22 = 9.577 mAs height of the wall is 7.31 m so the ball will clearthe wall.XtraEdge for IIT-JEE 29 MAY <strong>2011</strong>
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