A 3.18-kg projectile is fired with an initial speed of 122 m/s at an angle of 34

Question

A 3.18-kg projectile is fired with an initial speed of 122 m/s at an angle of 34° with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.99 kg and 2.19 kg. At 3.30 s after the explosion, the 2.19-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 0.99-kg fragment immediately after the explosion. (b) Find the distance between the point of firing and the point at which the 0.99-kg fragment strikes the ground. km (c) Determine the energy released in the explosion.

Explanation / Answer

at the heighest point, the projectile has only x component of velcoity.

vox = vo*cos(31)

= 122*cos(34)

= 101.1 m/s

height above the ground when explosion takes place, h = vo^2*sin^2(theta)/(2*g)

= 122^2*sin^2(34)/(2*9.8)

= 237.5 m

let M = 3.18 kg

m1 = 2.19 kg

m2 = 0.99 kg

let v1 and v2 are speed of m1 and m2 after the explosion.

v1x = 0 (beacuse it is mving down stright)

Apply, h = v1y*t + (1/2)*g*t^2

237.5 = v1y*3.3 + (1/2)*9.8*3.3^2

v1y = (237.5 - (1/2)*9.8*3.3^2)/3.3

= -55.8 m/s

Apply conservation of momentum in y-direction

0 = m1*v1y + m2*v2y

v2y = -m1*v1y/m2

= -2.19*(-55.8)/0.99

= 123.4 m/s

Apply conservation of momentum in x-direction.

M*vox = m1*0 + m2*v2x

v2x = M*vox/m2

= 3.18*101.1/0.99

= 324.7 m/s

so, v2 = v2xi + v2yj

= 324.7 (m/s) i + 123.4 (m/s) j <<<<<<<<<-------------Answer

let t is time taken for the second part to fall down.

Apply, -h = v2y*t - 0.5*g*t^2

-237.5 = 123.4*t - 4.9*t^2

4.9*t^2 - 123.4*t - 237.5 = 0

on sloving the above equation

we get

t = 26.98 m

distance travelled before landing = v2x*t

= 324.7*26.98

= 8760.4 m

distance travelled by first object = R/2

= vo^2*sin(2*theta)/(2*g)

= 122^2*sin(2*34)/(2*9.8)

= 704 m

so, the distance travelled by second fragmnet, = 704 + 8760

= 9464 m

= 9.46 km <<<<<<<<<<------------------Answer

c) at the top point, Ki = 0.5*M*vox^2

= 0.5*3.18*(101.1)^2

= 16252 J

Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*2.19*55.8^2 + 0.5*0.99*(324.7^2 + 123.4^2)

= 63135 J

Energy released in the explosion,

Kf - Ki = 63135 - 16252

= 46883 J

= 46.9 kJ <<<<<<<<<<------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.