Two particles, of masses M and 2M, moving, as shown, with speeds of 10m/s and 5m/s, collide elastically at the origin, After the collision, they move along the indicated directions with speeds v1 and v2, respectively. The values of v1 and v2 are nearly:
(a)
3.2m/s and 12.6m/s
(b)
3.2m/s and 6.3m/s
(c)
6.5m/s and 3.2 m/s
(d)
6.5m/s and 6.3m/s
(a)
3.2m/s and 12.6m/s
(b)
3.2m/s and 6.3m/s
(c)
6.5m/s and 3.2 m/s
(d)
6.5m/s and 6.3m/s
Question ID - 51335 | SaraNextGen Top Answer Two particles, of masses M and 2M, moving, as shown, with speeds of 10m/s and 5m/s, collide elastically at the origin, After the collision, they move along the indicated directions with speeds v1 and v2, respectively. The values of v1 and v2 are nearly:
(a)
3.2m/s and 12.6m/s
(b)
3.2m/s and 6.3m/s
(c)
6.5m/s and 3.2 m/s
(d)
6.5m/s and 6.3m/s
Two particles, of masses M and 2M, moving, as shown, with speeds of 10m/s and 5m/s, collide elastically at the origin, After the collision, they move along the indicated directions with speeds v1 and v2, respectively. The values of v1 and v2 are nearly:
|
(a) |
3.2m/s and 12.6m/s |
(b) |
3.2m/s and 6.3m/s |
|
(c) |
6.5m/s and 3.2 m/s |
(d) |
6.5m/s and 6.3m/s |
Apply conservation of linear momentum in X and Y direction for the system then
M(10cos300)+2M(5cos450)=2M(v1cos300)+M(v2cos450)
5
v1+
… (1)
Also 2M(5sin450)−M(10sin300)=2Mv1sin300−Mv2sin450
5
−5=v1−
… (2)
Solving equation (1) and (2)
(
+1)v1=5
+10
−5
⇒ v1=6.5m/s, v2=6.3m/s