A ball is thrown from the origin in the
x - y plane with velocity 28.28 m/s at an
angle 45° to the x - axis. At the same
instant a trolley also starts moving with
uniform velocity of 10m/s along the
positive x - axis. Initially, the trolley is
located
at 38m from the origin
Determine the time and position at
which the ball hits the trolley. height and widht of trolley is 2m and 3 m
Answers
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Explanation:
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Question from Class 11 Chapter Kinematics-2
On a frictionless horizontal surface , assumed to be the x−y plane , a small trolley A is moving along a straight line parallel to the y−aξs( see figure) with a constant velocity of (3–√−1)m/s . At a particular instant , when the line OA makes an angle of 45(∘) with the x−aξs , a ball is thrown along the surface from the origin O. Its velocity makes an angle ϕ with the x−aξsandithitsthetrol≤y.(a)Themotionoftheballisobservedomtheameofthetrol≤y.Calca−−tethe∠thetamadebythevelocity→→roftheballwiththe x-axis in this frame .
(b) Find the speed of the ball with respect to the surface , if ϕ=(4θ)/(4).
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Text Solution
Solution :
Method 1 : (a) Since the ball hits the trolley, relative to trolley, the velocity of ball should be directed towards the trolley. Hence, in the frame of trolley, the ball will appear to be moving towards OA, or in the frame of trolley, ball's velocity will make an angle of 45∘.
(b) ϕ=4θ3=4×45∘3=60∘
Using sine rule VBsin135∘=VAsin15∘
⇒VB=2ms−1
Method 2 : (a) Let A stands for trolley and B for ball.
Relative velocity of B will respect to A(v→BA) should be along OA for the ball to hit the trolley.
Hence, v→BA will make an angle of 45∘ with positive x - axis.
(b) tanθ=vBAyvBAx=tan45∘orvBAy=vBAx...(i)
Further vBAy=vBy−vAyorvBAx=vBx−0 ...(ii)
vBy−(3−1−−−−−√)
tanθ=vByvBxorvBy=vBxtanϕ
From (i),(ii),(iii), and (iv), we get
vBx=(3−1−−−−−√)tanϕ−1andvBy=(3−1−−−−−√)tanθ−1tanϕ
ϕ=4θ3=43(45∘)
Speed of ball w.r.t. surface,
vB=v2Bx+v2By−−−−−−−−√=3−1−−−−−√tanϕ−1secϕ
Substituting ϕ=60∘, we get vB=2ms−1.
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