Question

A bullet of mass M is fired with a velocity 50 m/s at an angle with the
horizontal. At the highest point of its trajectory, it collides head on
with a bob of mass 3M suspended by a massless rod of length
10
m and gets embedded in the bob. After the collision the rod
3
moves through an angle 120°. Find
[JEE-1998]
(a) The angle of projection.
(b) The vertical and horizontal coordinates of the initial position of the
bob with respect to the point of firing of the bullet. (g = 10m/s2)​

Answer 1

ANSWER

at height point velocity of mass m will be (50cosθ)ms

at height point velocity of mass m will be (50cosθ)ms at collision at highest point-

at height point velocity of mass m will be (50cosθ)ms at collision at highest point-consuming momentum

m(50cosθ)=4 mv¹

$m(50cosθ)=4 mv1</u></strong></p><p></p><p><strong><u>[tex]m(50cosθ)=4 mv1225cosθ=v1</u></strong></p><p></p><p><strong><u>[tex]m(50cosθ)=4 mv1225cosθ=v1$

now conserving energy (from hinge point of red)

charf kinetic energy = change in potential energy

$21(um)(v1)2=4 mg(L+dfrac22)$

$2m(v1)2⇒2(3)mgL$

$v2=3 gL  put y=10 L=310</u></strong></p><p><strong><u>[tex]v2=3 gL  put y=10 L=310$

$v1=10 m/s</u></strong></p><p></p><p><strong><u>[tex]v1=10 m/s \\ Put v1 in eq (1)</u></strong></p><p></p><p><strong><u>[tex]v1=10 m/s \\ Put v1 in eq (1)$

$25cosθ=10</u></strong></p><p></p><p><strong><u>[tex]25cosθ=10cosθ=54</u></strong></p><p></p><p></p><p><strong><u>[tex]25cosθ=10cosθ=54$

$cosθ=54</u></strong></p><p></p><p><strong><u>[tex]cosθ=54θ=cos−1(54)</u></strong></p><p></p><p><strong><u>[tex]cosθ=54θ=cos−1(54)θ=37o</u></strong></p><p></p><p><strong><u>[tex]cosθ=54θ=cos−1(54)θ=37o$

❤️

Answer 2

Answer:

Hope it helps dear

Explanation:

ANSWER

at height point velocity of mass m will be (50cos@ms

at height point velocity of mass m will be (50cosms at collision at highest point

at height point velocity of mass m will be (50cos)ms at collision at highest point

consuming momentum

m(50cos@)=4 my!

m(5000st) - 4 mel < /u >< /strong >< /p><p></P><p><strong><u> tex]m(50cos0) 4 mv1225cose v1 < /u>< /strong ></p><p>< /p><p> <strong><u> team(50cost) = 4 mv1225cost vl

now conserving energy (from hinge point of red)

charf kinetic energy = change in potential energy

21(um) (v1)2=4 mg(L+dfrac22)

2m(v1)2 = 2(3)mgl

12 = 3 gL put y = 10 L=310 < /u>< /strong></p><p><strong x u > [ter]v2 3 gL put y = 10 L=310

vl = 10 m/s < /u >< /strong >< /p><1 Put vi ineq (1) < /u >< /strong >< /p><

Pu

25cose = 10 < /u >< /strong >< /p>< p>< /p>< p>< strong ><u> [ter|25cose = 10cos0 = 54 < /u>< /strong >< /p><p>< /p><p></p> <p>< strong >< u> ter 25cost = 10cose = 54

cose = 54 < /u >< /strong >< /p><p> < /p><p>< strong ><u > [ter coso = 540 =006-1(54) </u></strong></p>