A is not possible because momentum is not conserved. A 1.2 kg spring-activated t

Question

A is not possible because momentum is not conserved. A 1.2 kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0.50 m/s. At the origin 0, the bomb explodes into two fragments. Fragment 1 has a mass of 0.4 kg and a speed of 0.9 m/s along the negative y-axis. In Figure 8.15, the x-component of the impulse on fragment 2, due to the explosion, is closest to: 0.6N-S 0.8N-S 0.4 N-s 0.2 N-s 1.0 N-s In Figure 8.17, four point masses are placed as shown. The x- and y-coordinates of the center of mass are closest to: (2.2,2.7) (2.3,2.8) (2.2, 2.6) (2.3, 2.6) (2.3,2.7)

Explanation / Answer

here,

mass of toy bomb , m = 1.2 kg

initial speed , u = 0.5 i m/s

final speed of m2 = 0.4 kg mass , v2 = - 0.9 m/s j

let the final speed of m1 = 0.8 kg be v1

using conseervation of momentum

m*u = m1*v1 + m2 * v2

1.2 * 0.5 i = 0.8 * v1 - 0.4 * 0.9 j

v1 = 0.75 i + 0.45 j

the x -component of the impulse , I = final momentum - initial momentum

I = 0.75 - 0.5 * 1.2

I = 0.15 kg.m/s

I = 0.2 kg.m/s

the x component of the impulse on fragment 2 is 0.2 kg.m/s

12)

the coordinates of the centre of mass ,

( x,y) = ( ( 8 *0 +4 * 3 + 5 * 6 +2 * 2 )/( 8 + 4 + 6 +2) , ( 8*3 + 5 *4 + 6*2 + 2*0) /( 8+4+6+2))

(x , y) = (2.3 m,2.8 m)

the x and y coordinate of the centre of mass are B) ( 2.3 m,2.8 m)

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