Question

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

Answer 1

@Vaidehi09

Answer 2

umm..is the answer -2.25 N-s ?

Answer 3

it should be 4.16 kg m/s

Answer 4

Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

Answer 5

Hw did u get that

Answer 6

http://gyazo.com/f78752c8ba71deacabb7a9450b8d2374

Answer 7

hw that 2 came

Answer 8

AO = Incident path
OB = Deflected path
∠AOB = Angle between the incident and deflected paths of the ball = 45°
∠AOP = ∠BOP = 22.5° = θ
got this?

Answer 9

isn't impulse = m(v2 - v1) ?

Answer 10

Now,according to the question,
Initial and final velocities of the ball = v right?
horizontal component of the initial velocity = vcos θ along RO
vertical component of the initial velocity = vsin θ along PO
horizontal component of the final velocity = vcos θ along OS
vertical component of the final velocity = vsin θ along OP

Answer 11

ok...thxxx

Answer 12

@Vaidehi09
∴Impulse imparted to the ball = Change in the linear momentum of the ball
so,mv cos theta = (-mvcos theta)
=>2mv cos theta

Answer 13

the horizontal components of velocities suffer no change the vertical components of velocities are in the opposite directions.

Answer 14

@DLS but what if we take the initial velocity as x-axis...then wouldn't the diag look like this? what's wrong with this?

Answer 15

if u take x axis